Answer:
y = -2/3x - 2
Step-by-step explanation:
Step 1: Find slope m of perpendicular line
Simply take the negative reciprocal of the given line
m = -2/3
y = -2/3x + b
Step 2: Find b
6 = -2/3(-12) + b
6 = 8 + b
b = -2
Step 3: Rewrite perpendicular equation
y = -2/3x - 2
What is the maximum value of the function f(x)=-x^2+6x+1 (enter an exact number) rotate image to see the problem
Answer:
10
Step-by-step explanation:
f(x)=-x^2+6x+1
This is a parabola that opens downward( the - coefficient of x^2)
The maximumx is at the vertex
The x coordinate is at
-b/2a where ax^2 + bx +c a =-1 b=6 c=1
-6/(2*-1)
-6/-2 = 3
The x coordinate of the vertex is 3
f(3) = - (3)^2 +6(3)=1
= -9+18+1
= 10
The vertex is ( 3,10)
The maximum value is 10
Answer:
[tex]10[/tex]
Step-by-step explanation:
[tex]f(x)=-x^2+6x+1[/tex]
x coordinate:
[tex]\frac{-b}{2a}[/tex]
[tex]a=-1\\b=6[/tex]
[tex]\frac{-6}{2(-1)} \\\frac{-6}{-2}\\ =3[/tex]
y-coordinate:
[tex]f(3)=-(3)^2+6(3)+1\\f(3)=-9+18+1\\f(3)=10[/tex]
What is the slope of the line through the points (2,8) and (5,7)
Answer:
-1/3
Step-by-step explanation:
The slope of the line can be found by
m = (y2-y1)/(x2-x1)
= ( 7-8)/(5-2)
= -1/3
Answer:
-1/3.
Step-by-step explanation:
The slope can be found by doing the rise over the run.
In this case, the rise is 8 - 7 = 1.
The run is 2 - 5 = -3.
So, the slope is 1 / -3 = -1/3.
Hope this helps!
PLEASE HELP!
What is the length of the shortest altitude in a triangle, if the lengths of the sides are 15 cm, 17 cm, 8 cm. NO DECIMALS
Answer:
The shortest altitude = 8 cm
Step-by-step explanation:
Where we have the sides given by
15 cm, 17 cm, 8 cm
From cosine rule, we have;
a² = b² + c² - 2×b×c×cos(A)
We have
For the side 15 cm,
15² = 17² + 8² - 2×17×8 cos A
-388 = -612×cos×A
A = 61.93°
17² = 15² + 8² - 2×15×8 ×cos B
0 = -240·cos B
B = 90°
Therefore, 17 is the hypotenuse side and 15 and 8 are the legs, either of which can be the height which gives the shortest altitude as 8 cm
An integer minus 5 times its reciprocal is
76
9
What is the integer?
Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him
Answer:
60 miles
Step-by-step explanation:
Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
Statement 1. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own
Statement 2. If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds
Solution
A. Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.
Let Brian speed=b
Ashok speed=a
Brian's walking speed=2(a-b)
b=2(a-b)
Divide both sides by 2
b/2=a-b
Ashok catches up in (time)= distance /( relative rate
=30/(a-b)
=30/(b/2)
=30÷b/2
=30*2/b
=60/b.
By that time Brian will cover a distance of
distance=rate*time
=b*60/b
=2(a-b)*60/2(a-b)
=60 miles
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds.
5a=3(a+b)
5a=3a+3b
5a-3a=3b
2a=3b
The baseball team has a double-header on Saturday. The probability that they will win both games is 50%. The probability that they will win just the first game is 65%, What is the probability that the team will win the 2nd game given that they have already won the first game? (PLEASE SHOW YOU'RE WORK)
Answer:
77%
Step-by-step explanation:
Given the following :
Probability of winning both games = 50%
Probability of winning just the first game = 65%
Let the probability of winning the ;
First game = p(A) = 65%
Second game = p(B)
Both games = p(A and B) = 50%
What is the probability that the team will win the 2nd game given that they have already won the first game
The above question is a conditional probability question :
Probability of winning the second Given that they've already won the first = p(B | A)
p(B | A) = (A and B) / p(A)
p(B | A) = 50% / 65%
p(B | A) = 0.5 / 0.65
p(B | A) = 0.7692307
= 76.9% = 77%
A particle moves along a straight line. The distance of the particle from the origin at time t is modeled by the equation below. s(t)equals2 sine t plus 3 cosine t Find a value of t between 0 and StartFraction pi Over 2 EndFraction that satisfies the equation s(t)equalsStartFraction 2 plus 3 StartRoot 3 EndRoot Over 2 EndFraction .
Answer:
The value of t that will satisfy the equation is π/6 (which is 30 degrees)
Step-by-step explanation:
The function that models the movement of the particle is given as;
S(t) = 2 sin(t) + 3 cos (t)
Now we want to the value of t between 0 and pi/2 that satisfies the equation;
s(t) = (2+ 3√3)/2 = 1 + 3√3/2
What we do here is simply find that value of t that would ensure that;
2sin(t) + 3cos(t) = 1 + 3√3/2
Without any need for rigorous calculations, this value of t can be gotten by inspection.
From our regular trigonometry, we know that the sin of angle 30 is 1/2 and its cos value is √3/2
We can make a substitution for it in this equation.
We obtain the following;
2 sin(30) + 3cos (30) and that is exactly equal to 1 + 3√3/2
Do not forget however that we have a range. And the range in question is between 0 and π/2
Kindly that π/2 in degrees is 90 degrees
So our range of values here is between 0 and 90 degrees.
So to follow the notation in the question, the value within the range that will satisfy the equation is π/6
PLEASE ANSWER THIS FAST Will the red square and the orange square always equal the blue square? What colored square is the hypotenuse?
Step-by-step explanation:
The area of the blue square will always equal the sum of the area of the orange and red rectanglesThe pythagorian theorem:
a²+b² = c²
now let a be the side of the red triangle and b the side of the orange one
so a² is the area of the red triangle and b² is the area of the orange one
Let c be the side of the blue rectangle
so c² is the area of it
then what we concluded is right
the hypotenuse is the blue side since it is the larger oneQuestion 4. In the graph, lines f and g intersect at P(6,6). What is the area, in square units, of the shaded region? * E. 15 F. 21 G. 27 H. 30
Answer:
E
Step-by-step explanation:
i guess the dotted lines outline a square
so get the area of the square which is 6×6=36
then don't focus on the shaded part but unshaded you'll see two right angled triangles
[tex]a = 1 \div2b \times h[/tex]
you will get a total for both as 21
then get the area of the square 36-21=15
so the area becomes 15
Simplify [tex]$\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$[/tex] $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$
Answer:
[tex]3 -\sqrt[2]3[/tex]
Step-by-step explanation:
Given
[tex]\frac{2\sqrt[3]{9}}{1 + \sqrt[3]{3} + \sqrt[3]{9}}[/tex]
Required
Simplify
Rewrite the given expression in index form
[tex]\frac{2 * 9 ^\frac{1}{3}}{1 + 3^{\frac{1}{3}} + 9^{\frac{1}{3}}}[/tex]
Express 9 as 3²
[tex]\frac{2 * 3^2^*^\frac{1}{3}}{1 + 3^{\frac{1}{3}} + 3^2^*^{\frac{1}{3}}}[/tex]
[tex]\frac{2 * 3^\frac{2}{3}}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}}}[/tex]
Multiply the numerator and denominator by [tex]1 - 3^{\frac{1}{3}}[/tex]
[tex]\frac{2 * 3^\frac{2}{3}}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}}} * \frac{1 - 3^{\frac{1}{3}}}{1 - 3^{\frac{1}{3}}}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) (1 - 3^{\frac{1}{3}})}{(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})(1 - 3^{\frac{1}{3}})}[/tex]
Open the bracket
[tex]\frac{2 (3^\frac{2}{3}) -2 (3^\frac{2}{3})(3^{\frac{1}{3}})}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}}(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})}[/tex]
Simplify the Numerator using Laws of Indices
[tex]\frac{2 (3^\frac{2}{3}) -2 (3^\frac{2+1}{3})}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}}(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})}[/tex]
Further Simplify
[tex]\frac{2 (3^\frac{2}{3}) -2 (3^\frac{3}{3})}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}}(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3^1)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}}(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}}(1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}})}[/tex]
Simplify the denominator
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}} - (3^{\frac{1}{3}})(3^{\frac{1}{3}}) - (3^{\frac{1}{3}})(3^{\frac{2}{3}})}[/tex]
Further Simplify Using Laws of Indices
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}} - (3^{\frac{1+1}{3}}) - (3^{\frac{1+2}{3}})}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}} - 3^{\frac{2}{3}} - 3^{\frac{3}{3}}}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}} - 3^{\frac{2}{3}} - 3^1}}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 + 3^{\frac{1}{3}} + 3^{\frac{2}{3}} - 3^{\frac{1}{3}} - 3^{\frac{2}{3}} - 3}}[/tex]
Collect Like Terms
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{1 - 3+ 3^{\frac{1}{3}} - 3^{\frac{1}{3}}+ 3^{\frac{2}{3}} - 3^{\frac{2}{3}} }}[/tex]
Group Like Terms for Clarity
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{(1 - 3) + (3^{\frac{1}{3}} - 3^{\frac{1}{3}}) + (3^{\frac{2}{3}} - 3^{\frac{2}{3}} )}}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{(- 2)+ (0) + (0)}}[/tex]
[tex]\frac{2 (3^\frac{2}{3}) -2 (3)}{-2}}[/tex]
Divide the fraction
[tex]-(3^\frac{2}{3}) + (3)[/tex]
Reorder the above expression
[tex]3 -3^\frac{2}{3}[/tex]
The expression can be represented as
[tex]3 -\sqrt[2]3[/tex]
Hence;
[tex]\frac{2\sqrt[3]{9}}{1 + \sqrt[3]{3} + \sqrt[3]{9}}[/tex] when simplified is equivalent to [tex]3 -\sqrt[2]3[/tex]
do these problems and get 100 points 1. Given the lengths of two sides of a triangle, find the range for the length of the third side. (Range means find between which two numbers the length of the third side must fall.) Write an inequality. c 22 and 15 2 Given the lengths of two sides of a triangle, find the range for the length of the third side. (Range means find between which two numbers the length of the third side must fall.) Write an inequality. d 13.2 and 6.7 3 Given the lengths of two sides of a triangle, find the range for the length of the third side. (Range means find between which two numbers the length of the third side must fall.) Write an inequality. e 34 and 12 4 Given the lengths of two sides of a triangle, find the range for the length of the third side. (Range means find between which two numbers the length of the third side must fall.) Write an inequality. f 23 and 44
Answer:
[tex]7 < x < 37[/tex] -- Triangle 1
[tex]6.5 < x < 19.9[/tex] -- Triangle 2
[tex]22 < x < 46[/tex] -- Triangle 3
[tex]21 < x < 67[/tex] -- Triangle 4
Step-by-Step Explanation:
Given
2 sides of a triangle
1. 22 and 15
2. 13.2 and 6.7
3. 34 and 12
4. 23 and 44
Required
Determine the range of the third side in the above triangles
Triangle 1: 22 and 15
Represent the third side with x
We'll make use of the following conditions to calculate the range of the third side;
[tex]22 + x > 15[/tex]
[tex]22 + 15 > x[/tex]
[tex]15 + x > 22[/tex]
Solving
[tex]22 + x > 15[/tex]
Make x the subject of formula
[tex]x > 15 - 22[/tex]
[tex]x > -7[/tex]
Solving
[tex]22 + 15 > x[/tex]
[tex]37 > x[/tex]
Solving
[tex]15 + x > 22[/tex]
Make x the subject of formula
[tex]x > 22 - 15[/tex]
[tex]x > 7[/tex]
The next step is to dismiss the inequality with negative digit; So, we're left with
[tex]37 > x[/tex] and [tex]x > 7[/tex]
Rewrite both inequalities
[tex]x < 37[/tex] and [tex]7 < x[/tex]
Combine the two inequalities
[tex]7 < x < 37[/tex]
Triangle 2: 13.2 and 6.7
Represent the third side with x
We'll make use of the following conditions to calculate the range of the third side;
[tex]13.2 + x > 6.7[/tex]
[tex]13.2 + 6.7 > x[/tex]
[tex]6.7 + x > 13.2[/tex]
Solving
[tex]13.2 + x > 6.7[/tex]
Make x the subject of formula
[tex]x > 6.7 - 13.2[/tex]
[tex]x > -6.5[/tex]
Solving
[tex]13.2 + 6.7 > x[/tex]
[tex]19.9 > x[/tex]
Solving
[tex]6.7 + x > 13.2[/tex]
Make x the subject of formula
[tex]x > 13.2 - 6.7[/tex]
[tex]x > 6.5[/tex]
The next step is to dismiss the inequality with negative digit; So, we're left with
[tex]19.9 > x[/tex] and [tex]x > 6.5[/tex]
Rewrite both inequalities
[tex]x < 19.9[/tex] and [tex]6.5 < x[/tex]
Combine the two inequalities
[tex]6.5 < x < 19.9[/tex]
Triangle 3: 34 and 12
Represent the third side with x
We'll make use of the following conditions to calculate the range of the third side;
[tex]34 + x > 12[/tex]
[tex]34 + 12 > x[/tex]
[tex]12 + x > 34[/tex]
Solving
[tex]34 + x > 12[/tex]
Make x the subject of formula
[tex]x > 12 - 34[/tex]
[tex]x > -22[/tex]
Solving
[tex]34 + 12 > x[/tex]
[tex]46 > x[/tex]
Solving
[tex]12 + x > 34[/tex]
Make x the subject of formula
[tex]x > 34 - 12[/tex]
[tex]x > 22[/tex]
The next step is to dismiss the inequality with negative digit; So, we're left with
[tex]46 > x[/tex] and [tex]x > 22[/tex]
Rewrite both inequalities
[tex]x < 46[/tex] and [tex]22 < x[/tex]
Combine the two inequalities
[tex]22 < x < 46[/tex]
Triangle 4: 23 and 44
Represent the third side with x
We'll make use of the following conditions to calculate the range of the third side;
[tex]23 + x > 44[/tex]
[tex]23 + 44 > x[/tex]
[tex]23 + x > 44[/tex]
Solving
[tex]23 + x > 44[/tex]
Make x the subject of formula
[tex]x > 23 - 44[/tex]
[tex]x > -21[/tex]
Solving
[tex]23 + 44 > x[/tex]
[tex]67 > x[/tex]
Solving
[tex]23 + x > 44[/tex]
Make x the subject of formula
[tex]x > 44 - 23[/tex]
[tex]x > 21[/tex]
The next step is to dismiss the inequality with negative digit; So, we're left with
[tex]67 > x[/tex] and [tex]x > 21[/tex]
Rewrite both inequalities
[tex]x < 67[/tex] and [tex]21 < x[/tex]
Combine the two inequalities
[tex]21 < x < 67[/tex]
simplify
[tex] {a}^{ - 2} {b}^{3} [/tex]
Answer:
Below
Step-by-step explanation:
● a^(-2) *b^3
●(1/a^2) *b^3
● b^3 / a^2
help me solve this Algebra problem please
Answer:
39858075
Step-by-step explanation:
Hello,
One basic way to see it is to compute the values.
75 = 25 * 3
225 = 75 * 3
675 = 225 * 3
etc ...
We can notice that this is multiplied by 3 every 10 years so we can compute as below.
year population
1970 25
1980 75
1990 225
2000 675
2010 2025
2020 6075
2030 18225
2040 54675
2050 164025
2060 492075
2070 1476225
2080 4428675
2090 13286025
2100 39858075
So the correct answer is 39858075
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
wht is the solution of the system defined be y =-x+5 and 5x+2y=14
Answer:
1.33,3.667
Step-by-step explanation:
Use y=mx+b for second system
which is y=5/2x+7
Now use substitution,graph, or elimination method.
PLEASE ASAP There are 20 players on a soccer team. From them, a captain and an alternate captain have to be chosen. How many possibilities are there?
Answer: 380
Step-by-step explanation:
Total number of players = 20
To select a captain and a vice captain :
One captain from a total of 20 players :
20 combination 1 = 20C1 = 20 ways
Total number of players left after selecting a captain = 20 - 1 = 19
Number of players left from which to select a vice captain = 19
Therefore, one vice captain from a group of 19 players :
19 combination 1 = 19C1 = 19 ways
Therefore no of different possible ways to select a captain and a vice captain :
20C1 × 19C1 = 20 × 19 = 380 Ways
Find the perimeter of a square with a diagonal of 15√2.
Answer:
15
Step-by-step explanation:
Answer:
21.213
Step-by-step explanation:
Jaden learns to perform 2 vocal pieces during each week of lessons . How many weeks of lessons will Jaden need before he will be able to sing a total of 24 pieces?
Answer:
12 weeks
Step-by-step explanation:
To solve this, all you need to do is divided 24 pieces by the two he learns per week. You'll then find it will take him 12 weeks
Which equation represents a line that is perpendicular to line FG? A. y=-1/2x+5 B. y=1/2x+2 C. y=-2x-3 D. y=2x-6
The equation of line which is perpendicular to the line FG is
y = -2x -3.
What is equation of line?
The equation of line is an algebraic form of representing the set of points, which together form a line in a coordinate system.
Formula for finding the equation of line from two points [tex](x_{1} ,y_{1} ) and (x_{2}, y_{2} )[/tex][tex](y -y_{1}) = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } (x-x_{1} )[/tex]
What is the slope of two perpendicular lines?If [tex]m_{1}[/tex] be the slope of one line, then the slope of the perpendicular line is [tex]\frac{-1}{m_{1} }[/tex].
What is the slope intercept form of a line ?The slope intercept form of the line is given by y = mx + b
Where, m is the slope of a line.
According to the given question
We have a line FG and the coordinates of points F and G are (-5,1) and (9,8) respectively.
Therefore, the slope of the line FG = [tex]\frac{8-1}{9+5}=\frac{7}{14} =\frac{1}{2}[/tex]
⇒ The slope of the line which is parallel to line FG is -2
Now, from the given option of the equation of line , y = -2x -3 has a slope of -2 .
Hence, the equation of line which is perpendicular to the line FG is
y = -2x -3.
Learn more about the equation of a perpendicular line here:
https://brainly.com/question/20712656
#SPJ2
A steel wire, when bent in the form of a square, encloses an area of 121 sq cm. The same wire
is bent, in the form of a circle. Find the area of the circle.
Answer:
A = 49[tex]\pi[/tex]
Step-by-step explanation:
First, we need to find the length of the wire. We can calculate this because we are given the area of the square, so we can work backwards.
Use the area formula and plug in the numbers:
A = s²
121 = s²
11 = s
We can calculate the length of the wire by multiplying 11 by 4, which is 44.
Now, we know the circumference of the circle is 44 units because that is how long the wire is.
We can work backwards again to find the radius, using the circumference formula:
C = 2[tex]\pi[/tex]r
44 = 2[tex]\pi[/tex]r
22 = [tex]\pi[/tex]r
7 = r
Now, we can find the area of the circle:
A = [tex]\pi[/tex]r²
A = [tex]\pi[/tex](7)²
A = 49[tex]\pi[/tex]
Which circle has a center angle that measures 40 degrees
Answer: B
Step-by-step explanation:
A center angle is an angle that has rays that originate from the center.
(05.05)
Based on the graph, what is the initial value of the linear relationship?
-4
-3
5/3
5
Answer:
5
Step-by-step explanation:
The initial value is when x=0
When x=0, y =5
The initial value is 5
Suppose you are interested in testing wheter the mean earning of men in the general social survey is representative of the earning of the entire U.S. Male population. If there are 372 men in the general social survey sample and approximately 128 million men in the population, calculate the degrees of freedom for this single-sample t test.
Answer:
371
Step-by-step explanation:
According to the given situation the calculation of degrees of freedom for this single-sample t test is shown below:-
Degrees of freedom is N - 1
Where N represents the number of Men
Now we will put the values into the above formula.
= 372 - 1
= 371
Therefore for calculating the degree of freedom we simply applied the above formula.
The graph for the equation y = x minus 4 is shown below. On a coordinate plane, a line goes through (0, negative 4) and (4, 0). Which equation, when graphed with the given equation, will form a system that has an infinite number of solutions? y minus x = negative 4 y minus x = negative 2 y minus 4 = x y + 4 x = 1 Brainliest reward
Answer:
The correct option is;
y minus x = negative 4
(y - x = -4)
Step-by-step explanation:
Given that the line y = x - 4 of the graph passes through the points (0, -4) and (4, 0)
Comparing with the general equation of a straight line, y = m·x + c, where m is the slope and c is the y-intercept gives;
The slope of the equation y = x - 4 = 1
The y-intercept of the equation y = x - 4 = -4
Two equations will have an infinite number of solutions when they are on the same line, that is having the same slope and intercept, we check for the slope and the intercept of the given options as follows;
For y - x = -4, we have;
y = x - 4 which is the same as the given equation and both equations will have an infinite number of solutions
For y - x = -2 we have;
y = x - 2
The slope is the same as the given equation but the intercept is different giving no solution
For y - 4 = x, we have;
y = x + 4
The slope is the same as the given equation but the intercept is different giving no solution
For y + 4x = 1, we have;
y = 1 - 4x
The slope and intercept are different giving one solution.
Answer:
y - x = -4
Step-by-step explanation:
Graph [tex]y=\frac{2}{3} x[/tex] Which of the following statements are true?
Answer:
A,C,D
Step-by-step explanation:
When b=0, there is a proportional relationship.
The slope in y=mx+b is the value next to x.
Using RISE/RUN when there is a change of 3 units in x, there is a change of 2 units in y.
Factor the polynomial expression x2 + 5.
Answer:
Step-by-step explanation:
Please write this as x^2 + 5.
Roots are ±i√5.
The corresponding factors of x^2 + 5 are (x + i√5) and (x - i√5)
Instructions: Find the measure of the indicated angle to the
nearest degree
Answer:
? = 23
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan ? = opp/ adj
tan ? = 3/7
take the inverse tan of each side
tan ^-1 tan ? = tan ^-1 ( 3/7)
? = 23.19859051
To the nearest degree
? = 23
The formula for the volume of a pyramid is =13ℎ
V
=
1
3
B
h
, where B is the area of the base and h is the height. Rearrange the formula to solve for the height (h).
Select one:
a. ℎ=3
h
=
3
V
B
b. ℎ=3
h
=
B
3
V
c. ℎ=3
h
=
V
3
B
d. ℎ=3
Answer:
h = V3B
Step-by-step explanation:
V = 1/3B · h
Divide volume by 1/3 B to get h by itself
V/1/3B = V3B
aljebra questions anum counted the money she had in her purse and money box . When she doubbled the amount in her purse and add it to the amount in money box she gets rs 1700 . When she triples the amount in her purse and add in to money box she gets rs 2200. How much money does she have in her purse
Answer:
Money in her purse is Rs. 500.
Step-by-step explanation:
Let the money in her purse = Rs. [tex]x\\[/tex]
Let the money in her Money box = Rs. [tex]y[/tex]
As per question statement,
Double the money in her purse (i.e. [tex]2 \times x[/tex]) and add it to the amount in money box, she gets Rs. 1700.[tex]2x+y=1700[/tex] ........ (1)
Triple the money in her purse (i.e. [tex]3\times x[/tex]) and add it to amount in money box ([tex]y[/tex]), she gets Rs. 2200.[tex]3x+y=2200[/tex] ....... (2)
To find: Money in her purse = ? i.e. [tex]x=?[/tex]
Let us solve for [tex]x[/tex] using the two linear equations.
We can use substitution method here i.e. find value of one variable from one equation and then substitute that value in other equation.
Using equation (1), we get the value of [tex]y[/tex] as follows:
[tex]y=1700-2x[/tex]
Now, let us put this value of y in equation (2) to find the value of [tex]x[/tex]:
[tex]3x+1700-2x=2200\\\Rightarrow x+1700 = 2200\\\Rightarrow x=2200-1700\\\Rightarrow x = Rs.\ 500[/tex]
Money in her purse is Rs. 500.
griffin ordered a pair of sneakers online. he had 16 credit that he applied toward the purchase, and then he used a credit card to pay for the rest of the cost. if the the shoes cost 80, then how much did griffin charge to his credit card when he bought the sneakers? PLEASE ANSWER I BEG Y'ALL
Answer: Griffin charge $79.854 to his credit card when he bought the sneakers.
Step-by-step explanation:
Griffin ordered a pair of sneakers online.
Value of each credit point = 1 cent
Then , value of 16 credit points = 16 cents = $0.16 [1$ = 100 cents]
Cost of shoes = Rs $80
Charge to credit card = (Cost of shoes) - (Value of 16 credit points)
= $(80-0.16)
= $79.84
Hence, Griffin charge $79.854 to his credit card when he bought the sneakers.
Hi May I know how to solve this step by step please
Answer:
2, 3 , 5, 7
Step-by-step explanation:
2(x - 2)/3 < (x + 1)/2 < 3(5x + 6)/4
Considering:
2(x - 2)/3 < (x + 1)/2
<=>(2x - 4)/3 < (x + 1)/2
<=> (2x - 4)*2 < (x + 1)*3
<=> 4x - 8 < 3x + 3
<=> 4x - 3x < 8 + 3
<=> x < 11
Considering:
(x + 1)/2 < 3(5x + 6)/4
<=>(x + 1)/2 < (15x + 18)/4
<=>(x + 1)*4 < (15x + 18)*2
<=> 4x + 4 < 30x + 36
<=> 4x - 30x < 36 - 4
<=> -26x < 32
<=> 26x > -32
<=> x > -32/26
=> -32/26 < x < 11
The prime numbers satisfy the above inequalities: 2, 3 , 5, 7