Step-by-step explanation:
13
[tex]3 + (20 \div 5) - ( {2}^{2}) [/tex]
[tex]3 + 4 - 4 = 3[/tex]
15
[tex] \frac{10 - {3}^{2} }{20 - (3 \times 4)} [/tex]
[tex] \frac{10 - 9}{20 - 12} [/tex]
[tex] \frac{1}{8} [/tex]
A motorboat travels 116 kilometers in 4 hours going upstream. It travels 164 kilometers going downstream in the same amount of time. What is the rate of the
boat in still water and what is the rate of the current?
Rate of the boat in still water:
Rate of the current:
9514 1404 393
Answer:
boat: 35 km/hcurrent: 6 km/hStep-by-step explanation:
Let b and c represent the rates of the boat and current, respectively. Then the upstream rate is ...
speed = distance/time
b -c = (116 km)/(4 h) = 29 km/h
b +c = (164 km)/(4 h) = 41 km/h
Adding these equations gives ...
2b = 70 km/h
b = 35 km/h
Substituting into the first equation, we have ...
35 -c = 29
c = 6
Rate of the boat: 35 km/h.
Rate of the current: 6 km/h
What is the solution to the system of equations below?
4x-2y=10
y=-2x-5
show your work! :)
Answer:
(0,-5)
Step-by-step explanation:
4x-2(-2x-5)+10
x=0
y=-2*0-5
y=-5
how many inches is 10560 feet?
Answer:
126720
Step-by-step explanation:
Pre-algbra worksheet!!Please help!! will give brainliest!!
Answer:
1/4= 0.5=0.125, 2/4=0.5=0.125, 3/4= 0.75=0.1878
pls help im a little bit sтuрid
Step-by-step explanation:
a⁴+16a²b-2ab-a²+64b²-b²we can also write them as,
a⁴+16a²b+64b²-a²-2ab-b²Now, from -a²-2ab-b² if we take - as common term then we get,
a⁴+16a²b+64b²-(a²+2ab+b²)Now, by using the concept of a²+2ab+b²=(a+b)²
(a²)²+2×a²×8b+(8b)²-{(a)²+2×a×b+(b)²}(a²+8b)²-(a+b)²I hope it helped U
stay safe stay happy
Please help me solve that question
Step-by-step explanation:
steps are in picture above.
if you need to ask questions I'll give you answers gladly.
Use the graph to find out over what interval f(x) is less than or equal to 0. PLEASE HELP ASAP!!!
Answer:
[-5, 2]
Step-by-step explanation:
We have to find the interval for which f(x) <= 0, the interval is [-5, 2].
Solve the following problems using equations. Show your working steps clearly.
1. Four years ago, the age of Paul was twice that of John. John is 18 years old now. How old is Paul now?
2. Five bottles of milk cost $4.4 more than four bottles of juice. If one bottle of milk costs $3.2, find the cost of one bottle of juice.
3. Miss Cheung spends 1/8 of her savings on personal ornaments and 5/6 on clothing. If her savings still has $130 left, find the original amount of her savings.
1.
Paul's present age = x
John's present age = 18
Paul's age 4 years ago = x - 4
John's age 4 years ago = 18 - 4 = 14
Also,
Paul's age four years ago was twice that of John's age four years ago.
Therefore,
Paul's age 4 years ago = 2(14)
= 28
Paul's age 4 years ago = x - 4 = 28
x - 4 = 28
x = 28 + 4
x = 32
Therefore, the present age of Paul is 32.
2.
Cost of one bottle of milk = $3.2
Cost of five bottles of milk = $3.2 × 5 = $15
Cost of four bottles of juice = $15 - $4.4
= $10.6
Cost of one bottle of juice = x
= Cost of four bottles of juice
4
= 10.6
4
= 2.56
Therefore, one bottle of juice costs $2.56.
3.
Savings spent on personal ornaments = 1/8
Savings spent on clothing = 5/6
Total savings spent = 1/8 + 5/6
= 3/24 + 20/24
= (3+20)/24
= 23/24
Savings left = 1/24 = $130
Total amount of savings = x
As 130 is one part of the total savings,
Total amount of savings =
x = 130 × 24
x = 3120
Therefore, original amount of savings is $3120.
PLEASE HELP
1= a-13/-6
Show your work in details if you can, I have a hard time understanding this.
Answer: 7
Step-by-step explanation:
1. multiply both sides by -6 so the -6 on the right side could cancel out and the left side would be -6: 1= a-13/-6 ==> -6=a-13
2. add both sides by 13 so -13 on the right side could cancel out and the left side would be -6+13: -6=a-13 ==> -6+13=a-13+13
3. you know -6+13 equals 7 right? so the answer is a=7
Hey there! I'm happy to help!
We want to get the a all by itself. For me, I always like to have the variable on the left side, so first, let's just flip this around.
[tex]\frac{a-13}{-6} =1[/tex]
What we want to do is get all of the constants (numbers without variables) on the right side. This will leave a all on its own.
We see that the quantity a-13 is being divided by -6. To undo this, we will just multiply both sides by -6. This will cancel out that -6 and get us closer to having the a isolated.
[tex]\frac{a-13}{-6} *-6=1*-6\\\frac{a-13}{1}=-6\\a-13=-6[/tex]
Now we add 13 to both sides to undo the -13 on the left side.
[tex]a-13+13=-6+13[/tex]
[tex]a+0=7\\a=7[/tex]
Have a wonderful day and keep on learning! :D
For the two numbers listed, find two factors of the first number such that their product is the first number and their sum is the second number.
- 10,9
9514 1404 393
Answer:
10, -1
Step-by-step explanation:
If you assume the numbers are integers, you can look at the factor pairs that make -10:
-10 = 10(-1) = 5(-2) = 2(-5) = 1(-10)
The sums of these pairs are 9, 3, -3, -9, so the pair of interest is the first one:
10, -1
Please help me It’s my birthday!
2. Janyce conjectured that in the visual pattern below, there would be 3n – 1 small
squares in the nth figure
Can you disprove Janyce’s conjecture?
Step-by-step explanation:
Janyce's rule doesn't work. In the case where n = 3 her rule predicts that there will be 3(3) - 1 = 8 squares but instead, there are 10 squares. Likewise, for n = 4, there are 17 squares instead of 11 that she predicted. Instead, I propose the rule [tex]n^2 + 1.[/tex] It works for all the figures shown and mostly likely, for all other values of n.
2. Part A;
Given that the number of squares obtained using Janyce conjecture for
figures 3 and 4 which are 8 and 11 squares respectively and the number of
squares in the diagram for figures 3 and 4 which are 10 and 17 squares
respectively are not the same, disproves Janyce conjecture
Part B:
The rule for the number of squares in the nth squares is n² + 1
The reasons the above conclusion and rule are correct is as follows:
2. Part A
The given number of squares in each figure are;
[tex]\begin{array}{|c|c|c|}\mathbf{Figure \ number}&\mathbf{Number \ of \ squares} &\mathbf{Janyce \ conjecture \ (3\cdot n - 1)}\\&&\\1&2 \ squares &(3 \times 1 - 1 = 2) \ squares \\2&5 \ squares&(3 \times 2 - 1 = 5) \ squares\\3&10 \ squares &(3 \times 3 - 1 = 8) \ squares\\4&17 \ squares&(3 \times 4 - 1 = 11) \ squares\end{array}[/tex]
Given that the number squares in figure 3, and 4 are;
Figure 3 = 10 squares
Figure 4 = 17 squares
Janyce conjectured number of squares are;
Figure 3 = 8 squares
Figure 4 = 11 squares
The values for the number of squares in the diagram in figures 3 and 4 are different from Janyce conjecture which disproves Janyce's conjecture for the number of squares in the nth figure
Part B
From the diagrams, we have;
The number of squares on the each figure is given by a square shape with a side length equal to the figure number of squares plus one extra square at the top right end corner, we have;
The area of the square part of the shape = n² = The number of squares in the square shape
The number of squares in the nth figure = n² + 1
Verifying, we have;
Figure 1 = (1² + 1) squares = 2 squares (Same as in diagram)
Figure 2 = (2² + 1) squares = 5 squares (Same as in diagram)
Figure 3 = (3² + 1) squares = 10 squares (Same as in diagram)
Figure 4 = (4² + 1) squares = 17 squares (Same as in diagram)
Therefore;
The correct formula for the number of squares in the nth figure = n² + 1
Learn more about the nth term of a sequence here:
https://brainly.com/question/12998719
https://brainly.com/question/15325765
https://brainly.com/question/21493252
Let the (x; y) coordinates represent locations on the ground. The height h of
a particle is governed by the function described below. Classify all the critical
points of height and find the value(s) of height h at those points.
h(x; y) = 8x + 10y - 4xy - 4x^2 - 4y^3
The critical points of h(x,y) occur wherever its partial derivatives [tex]h_x[/tex] and [tex]h_y[/tex] vanish simultaneously. We have
[tex]h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5[/tex]
Substitute y in the second equation and solve for x, then for y :
[tex]2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}[/tex]
This is to say there are two critical points,
[tex](x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)[/tex]
To classify these critical points, we carry out the second partial derivative test. h(x,y) has Hessian
[tex]H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}[/tex]
whose determinant is [tex]192y-16[/tex]. Now,
• if the Hessian determinant is negative at a given critical point, then you have a saddle point
• if both the determinant and [tex]h_{xx}[/tex] are positive at the point, then it's a local minimum
• if the determinant is positive and [tex]h_{xx}[/tex] is negative, then it's a local maximum
• otherwise the test fails
We have
[tex]\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0[/tex]
while
[tex]\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0[/tex]
So, we end up with
[tex]h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}[/tex]
Let C be the positively oriented square with vertices (0,0), (3,0), (3,3), (0,3). Use Green's Theorem to evaluate the line integral
∫C 8y^2 x dx + 6x^2 y dy
Let D denote the region whose boundary is C. By Green's theorem,
[tex]\displaystyle \int_C 8y^2x\,\mathrm dx + 6x^2y\,\mathrm dy = \iint_D \frac{\partial(6x^2y)}{\partial x} - \frac{\partial(8y^2x)}{\partial y} \,\mathrm dx\,\mathrm dy \\\\ = \int_0^3 \int_0^3 (12xy-16yx)\,\mathrm dx\,\mathrm dy \\\\ = -4 \int_0^3 \int_0^3 xy\,\mathrm dx\,\mathrm dy = \boxed{-81}[/tex]
x=-9 y=-8 evaluate the expression 8 x^2 - 6 y^2
Answer:
264Step-by-step explanation:
Let x= -9 y= -88x² - 6y² = ?8(-9)² - 6(-8)² = ?8(81) - 6(64) = ?648 - 384 = ?264[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]
9(y-4) -3y = 2(3y-2 need solved as fast as possible
Step-by-step explanation:
did you give us all the information ? the equating looks cut off.
so, from what I can see we have
9(y-4) - 3y = 2(3y-2)
let's try :
9y - 36 - 3y = 6y - 4
0 = 32
so, no, this can't be it. this does not have a solution. you must have cut off some additional information.
how many feet is 2160 yards?
In the number 6.253
(a) What is the place value of 2 ...................................................................
(b) What is the value represented by 5 ...................................................................
Answer:
place value of 2 is 100
Step-by-step explanation:
please brainleist my answer
Answer:
The place value of 2 in 6.253 is 200
Step-by-step explanation:
Place value of 2 is occurring in hundreds
8x – 3 = 11 + 6x
What is x
Answer:
x = 7
Step-by-step explanation:
[tex]8x-3=11+6x\\\\2x-3=11\\\\2x=14\\\\x=7[/tex]
Answer:
x = 7
Step-by-step explanation:
8x - 3 = 11 + 6x
8x - 3 + 3 = 11 + 6x +3
8x = 6x + 14
8x - 6x = 6x + 14 - 6x
2x = 14
2x/2 = 14/2
x = 7
Please solve the equation for z
Answer:
z = 2y - 4x
Step-by-step explanation:
[tex]{ \sf{x = \frac{2y - z}{4} }} [/tex]
multiply both sides by 4:
[tex]{ \sf{(4 \times x) = ( \frac{2y - z}{4}) \times 4 }} \\ \\ { \sf{4x = 2y - z}} \\{ \sf{z = 2y - 4x}}[/tex]
create a improper fraction with the number 10 as the denominator
Answer:
Decimal Fraction Percentage
0.101 101/1000 10.1%
0.1 100/1000 10%
0.1013 101/997 10.13%
0.1012 101/998 10.12%
Step-by-step explanation:
Calculate the area of each figure.
Answer:
230cm²
Step-by-step explanation:
We can split the shape into 2 parts - the triangle and the rectangle.
The area of the retangle is 10*17cm = 170cm²
The area of the triangle = 1/2 * (22-10) * (17-7)
= 1/2 * 12 * 10 = 60
Therefore the total area is equal to 170+60 = 230cm²
Evaluate: 24 - ( 5 - 3)^ 4
______________
4
24 - (5 - 3)⁴ = 24 - 2⁴ = 24 - 16 = 8
8 : 4 = 2 ← the end solution
2. The product of 7 and a number
I think the answer you are looking for is 7n, which also means 7 times n.
Answer:
7x
Step-by-step explanation:
Product refers to multiplication. "A number" refers to an unknown number otherwise known as a variable. In this case, I use x.
help me plz branniest to whoever right
a bakery sells 17 cups of coffee for every 3 cups of hot chocolate they sell. At the end of the day the bakery sold 406 more coffee than hot chocolates. How many of the two drinks did they sell in all?
Answer:
Step-by-step explanation:
The cost of petrol is 60 rupees per litre a petrol bunk sells 750
litre s of petrol one day. how much money do they get At The end of the day
Answer:
45000rupees
Step-by-step explanation:
cost of 1 litre of petrol=60rupees
cost of 750 litres of petrol=750×60rupees
=45000rupees
What is the range with steps? f(x)=x^2+8x+8
Answer:
[-8, ∞)Step-by-step explanation:
Given function:
f(x)=x² + 8x + 8This is a quadratic function with positive leading coefficient, therefore it is an increasing function with its minimum at vertex. It has no maximum value.
The vertex is at x = -b/2a:
x = -8/2 = -4The minimum value is:
f(-4) = (-4)² + 8(-4) + 8 = 16 - 32 + 8 = -8The range of the function is:
[-8, ∞)f(x)=x^2+8x+8
Find vertex to x[tex]\\ \sf\longmapsto \dfrac{-b}{2a}[/tex]
[tex]\\ \sf\longmapsto \dfrac{-8}{2(1)}[/tex]
[tex]\\ \sf\longmapsto \dfrac{-8}{2}[/tex]
[tex]\\ \sf\longmapsto -4[/tex]
Find value of function at -4[tex]\\ \sf\longmapsto f(-4)[/tex]
[tex]\\ \sf\longmapsto (-4)^2+8(-4)+8[/tex]
[tex]\\ \sf\longmapsto 16-32+8[/tex]
[tex]\\ \sf\longmapsto -16+8[/tex]
[tex]\\ \sf\longmapsto -8[/tex]
The range is
[tex]\\ \sf\longmapsto (8,\infty)[/tex]
Ayudaa con esa qnq
-9/4 + 3/7 x -1/11
Answer:
3x-721/44 over 7
Step-by-step explanation:
1)Combine multiplied terms into a single fraction
2)Multiply the numbers
3)Subtract the numbers
4)Find common denominator
5)Combine fractions with common denominator
6)Multiply the numbers
7)Rearrange terms
You move left 4 units. You end at (-5, -5). Where did you start?
Answer:
(-1,-5)
Step-by-step explanation:
To understand where this point was before the transformation you must first understand the transformation itself. Shifting left is a horizontal translation, this means that it will affect the x-coordinate. The x-coordinate is the first digit in the pair. Therefore, in the case of (3,5), 3 would be the x-coordinate. Additionally, moving left is moving towards the negative, so it is the same as subtracting. Thus, to find where you started, work backward. Do this by adding 4 to the x-coordinate. This means the beginning coordinate was (-1,-5).
Factor x^2+x+6
Plz help
Answer:
2
Step-by-step explanation: