how to do this question plz answer ​

Answer:

2

Step-by-step explanation:

green

Answer: 2

Step-by-step explanation:

Answer:

675 – (15-12) ³+3

675-3³ +3

675-27+3

651

Step-by-step explanation:

The value of the given expression is 666 which is the correct answer that would be an option (C).

PEMDAS rule states that the order of operation starts with the calculation enclosed in brackets or the parentheses first. Exponents (degrees or square roots) are then operated on, followed by multiplication and division operations, and then addition and subtraction.

We have been given the expression below as:

675 – (15-12)³ ÷ 3

Using the PEMDAS rule to determine the evaluation of the given expression  

⇒ 675 – (15-12)³ ÷ 3

Apply the subtraction operation,

⇒ 675 - (3)³ ÷ 3

⇒ 675 - 27 ÷ 3

⇒ 675 - 27 / 3

Apply the division operation,

⇒ 675 - 9

Apply the subtraction operation,

⇒ 666

Therefore, the value of the given expression would be 666,

Learn more about the PEMDAS rule here :

https://brainly.com/question/20876480

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Answer:

b=-a

Step-by-step explanation:

Answer:

'll tell you where the problem lies - it is IMPOSSIBLE to form triangles like this.

If the perimeter of the smallest triangle is 6 and one side is 3, then the sum of the other two sides can only be 6 - 3 = 3

One property to enable you to form a triangle is that NO ONE SIDE can be greater or equal to the sum of the other two sides. In the smallest triangle 1 side of length 3 equals the other two sides.

In the middle triangle one side of length "4" equals the sum of the other two sides and

In the large triangle one side of length "5" equals the other two sides.

Therefore when I say "triangle" above I am not actually correct because it is IMPOSSIBLE to form triangles with those dimensions of 1 side and with those perimeters

Answer:

After 32 minutes the two pools will have the same amount of water.

There will be 1,366 liters in each pool when they have the same amount of water.

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations:

The first pool contains 720 liters of water, and it’s being added at a rate of 19.25 liters per minute.

First Pool = 720 +19.25 m

Where m is the number of minutes.

The second pool is empty, and Water is being added at a rate of 41.75 liters per minute

Second pool = 41.75m

Since both pools must have the same amount of water:

720 +19.25 m = 41.75m

Solving for m:

720 = 41.75m-19.25 m

720 = 22.5m

720/22.5 = m

32 = m

After 32 minutes the two pools will have the same amount of water.

Finally, we replace m=32 on any equation:

41.75m = 41.75 (32) = 1,336 liters

There will be 1,366 liters in each pool when they have the same amount of water.

Feel free to ask for more if needed or if you did not understand something.  

Answer:

Simply subtract the sum of the the three angles given from 360° in order to get the measure of the fourth angle!

Step-by-step explanation:

Answer:

The probability that the first card is a Heart and the second card is a Spade is 0.064.

Step-by-step explanation:

A standard deck of 52 cards is shuffled and two cards are drawn without replacement.

The denominations of the cards are as follows:

Spades (S) = 13

Hearts (H) = 13

Diamonds (D) = 13

Clubs (C) = 13

Compute the probability of selecting a Heart first as follows:

[tex]P(H)=\frac{13}{52}=0.25[/tex]

Compute the probability of selecting a Spade second as follows:

[tex]P(S)=\frac{13}{51}=0.255[/tex]

Since the two cards are selected without replacement the second draw is independent of the other.

Then the probability that the first card is a Heart and the second card is a Spade is:

[tex]P(1st\ H\cap 2nd\ S)=P(H)\times P(S)[/tex]

                            [tex]=0.25\times 0.255\\=0.06375\\\approx 0.064[/tex]

Thus, the probability that the first card is a Heart and the second card is a Spade is 0.064.

Answer:

a) [tex]\boxed{4m^2+20m+25}[/tex]

b) [tex]\boxed{m^2-m+\frac{1}{4} }[/tex]

Step-by-step explanation:

a) [tex](2m+5)^2[/tex]

Using formula [tex](a+b)^2 = a^2+2ab+b^2[/tex]

=> [tex](2m)^2 + 2(2m)(5)+(5)^2[/tex]

=> [tex]4m^2+20m+25[/tex]

b) [tex](m-\frac{1}{2} )^2[/tex]

Using Formula [tex](a-b)^2 = a^2-2ab+b^2[/tex]

=> [tex](m)^2 - 2(m)(\frac{1}{2} ) + (\frac{1}{2} )^2[/tex]

=> [tex]m^2-m+\frac{1}{4}[/tex]

Answer:

a) [tex]\boxed{4m^2 + 20m + 25}[/tex]

b) [tex]\boxed{m^2 - m + \frac{1}{4} }[/tex]

Step-by-step explanation:

[tex](2m + 5)^2[/tex]

[tex](2m + 5) (2m + 5)[/tex]

Use FOIL method.

[tex]2m(2m + 5)+5(2m + 5)[/tex]

[tex]4m^2 + 10m + 10m + 25[/tex]

[tex]4m^2 + 20m + 25[/tex]

[tex](m - \frac{1}{2} )^2[/tex]

[tex](m- \frac{1}{2})(m- \frac{1}{2})[/tex]

Use FOIL method.

[tex]m(m- \frac{1}{2})- \frac{1}{2}(m- \frac{1}{2})[/tex]

[tex]m^2- \frac{1}{2} m- \frac{1}{2}m+ \frac{1}{4}[/tex]

[tex]m^2-m+ \frac{1}{4}[/tex]

Answer: 320 students

Step-by-step explanation:

From the question, we are informed that a school counselor surveyed 90 randomly selected students about thé langages they speak and thé students surveyed 16 speak more than one langage fluently. This means that 16/90 speak more than one language.

When 1800 students are surveyed, the number of students that can be expected to speak more than one langage fluently will be:

= 16/90 × 1800

= 16 × 20

= 320 students

Answer:

A.) Mean = 237.9

B.) Median = 240

C.) Mode = 199

D.) Midrange = 239.5

Step-by-step explanation:

The given data are :

293 255 264 240 190 295 199 184 293 205 199

The mean = (sum of X) / f

Where frequency f = 11

X = 293 + 255 + 264 + 240 + 190 + 295 + 199 + 184 + 293 + 205 + 199

X = 2617

Substitute X and f into the formula

Mean = 2617/11

Mean = 237.9 approximately

B.) To get the median, you need to first rearrange the data, then pick the middle number.

184 190 199 199 205 240 255 264 293 293 295

The median = 240

C.) The mode is the highest frequency. That is the most occuring number

Mode = the two most occuring numbers are 199 and 293

D.) Range = highest number - lowest number

But midrange = (highest number + lowest number ) ÷ 2

Highest number = 295

Lowest number = 184

Substitute into the formula

Midrange = (295 + 184)/2

Midrange = 479/2

Midrange = 239.5

Answer:

Katie is correct. You would take 20% of $22.50 (22.5 multiplied by .2). You would get $4.50 off of the book with the discount. So you would subtract 4.5 from 22.5 and get $18. Then you would take 10% of $18 for the sales tax. (18 multiplied by .1). You would get $1.80 towards sales tax. you would then add $1.80 to $18 and get $19.80

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Answer:

Approximatley 5.8 units.

Step-by-step explanation:

We are given two angles, ∠S and ∠T, and the side opposite to ∠T. We need to find the unknown side opposite to ∠S. Therefore, we can use the Law of Sines. The Law of Sines states that:

[tex]\frac{\sin(A)}{a}=\frac{\sin(B)}{b} =\frac{\sin(C)}{c}[/tex]

Replacing them with the respective variables, we have:

[tex]\frac{\sin(S)}{s} =\frac{\sin(T)}{t} =\frac{\sin(R)}{r}[/tex]

Plug in what we know. 20° for ∠S, 17° for ∠T, and 5 for t. Ignore the third term:

[tex]\frac{\sin(20)}{s}=\frac{\\sin(17)}{5}[/tex]

Solve for s, the unknown side. Cross multiply:

[tex]\frac{\sin(20)}{s}=\frac{\sin(17)}{5}\\5\sin(20)=s\sin(17)\\s=\frac{5\sin(20)}{\sin(17)} \\s\approx5.8491\approx5.8[/tex]

Answer:

Option (4) and Option (5)

Step-by-step explanation:

By calculating the second difference, if the second difference in a table is equal, table will represent the quadratic relationship.

In the given option, we analyze that table given in Option (4) will represent the quadratic relationship.

x             y             Ist difference [tex](y_2-y_1)[/tex]         IInd difference

0            4                      -                                             -

1            -4             -4 - (4) = -8                                     -    

2           -4             -4 - (-4) = 0                              0 - (-8) = 8

3            4              4 - (-4) = 8                                  8 - 0 = 8

Second difference of the terms in y are the same as 8.

Therefore, table of Option (4) represents the quadratic relationship.

Similarly, in Option (5) we will calculate the second difference of y terms.

x            y            Ist difference     IInd difference

0          -4                    -                            -

1           -8             -8 - (-4) = -4                 -

2          -10           -10 - (-8) = -2        -2 - (-4) = 2      

3          -10           -10 - (-10) = 0        0 - (-2) = 2

Here the second difference is same as 2.

Therefore, table of Option (5) will represent the quadratic relationship.

Answer:

Option 5 is wrong

Step-by-step explanation:

Answer:

The number is 3

Step-by-step explanation:

The fraction is 11/36

let

x = no. added to the numerator

2x = no. added to denominator

We have,

x+11/2x+36=1/3

Cross multiply

3(x+11) = (2x + 36)

3x + 33 = 2x + 36

Collect like terms

3x - 2x = 36 - 33

x=3

The number is 3

Check:

3+11/6+36=1/3

14/42=1/3

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  [tex]$y'(t)=kb2t\cos(bt^2)$[/tex]

(c).  [tex]$ b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$[/tex]

Step-by-step explanation:

(a). Let the curve is,

[tex]$y(t)=k \sin (bt^2)$[/tex]

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value [tex]x_{0}[/tex]  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    [tex]$y(t)=k \sin (bt^2)$[/tex]

Differentiating the above equation with respect to x, we get

[tex]y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)][/tex]

Applying chain rule,

[tex]y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)[/tex]  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

[tex]$y(t)=k \sin (bt^2)$[/tex]

[tex]$y'(t)=kb2t\cos (bt^2)$[/tex]

Now putting the initial conditions in the equation y'(1)=0

[tex]$kb2(1)\cos(b(1)^2)=0$[/tex]

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

[tex]$b=\frac{\pi}{2}$[/tex]

And

y'(2) = 3

[tex]$\therefore kb2(2)\cos [b(2)^2]=3$[/tex]

[tex]$4kb\cos (4b)=3$[/tex]

[tex]$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$[/tex]

[tex]$2k\pi\cos 2 \pi=3$[/tex]

[tex]2k\pi(1) = 3$[/tex]  

[tex]$k=\frac{3}{2\pi}$[/tex]

[tex]$\therefore b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$[/tex]

The y'(1) =0, y'(2) = 3, and the  [tex]\rm y'(t) = kb2t \ cos(bt^2)[/tex]  and value of b and k are [tex]\pi/2[/tex]  and  [tex]3/2\pi[/tex] respectively.

It is given that the curve  [tex]\rm y(t) = ksin(bt^2)[/tex]

It is required to find the critical point, first derivative, and smallest value of b.

It is defined as a special type of relationship and they have a predefined domain and range according to the function.

We have a curve:

[tex]\rm y(t) = ksin(bt^2)[/tex]

Given that the first critical point of y(t) for positive t occurs at t = 1

First, we have to find the first derivative of the function or curve:

[tex]\rm y'(t) = \frac{d}{dt} (ksin(bt^2))[/tex]

[tex]\rm y'(t) = k\times2bt\times cos(bt^2)[/tex]   [ using chain rule]

[tex]\rm y'(t) = kb2t \ cos(bt^2)[/tex]

y(0) = 0

y'(0) = 0

The critical point is the point where the derivative of the function becomes 0 at that point in the domain of a function.

From the critical point y'(1) = 0 ⇒  [tex]\rm kb2 \ cos(b) =0[/tex]

k and b can not be zero

[tex]\rm cos(b) = 0[/tex]

b = [tex]\rm \frac{\pi}{2}[/tex]

and y'(2) =3

[tex]\rm y'(2) = kb2\times 2 \times cos(b\times2^2) =3\\\\\rm 4kb \ cos(4b) =3[/tex](b =[tex]\rm \frac{\pi}{2}[/tex])

[tex]\rm 4k\frac{\pi}{2} \ cos(4\frac{\pi}{2} ) =3\\\\\rm2 \pi kcos(2\pi) = 3[/tex]

[tex]\rm2 \pi k\times1) = 3\\\rm k = \frac{3}{2\pi}[/tex]

Thus, y'(1) =0, y'(2) = 3, and the  [tex]\rm y'(t) = kb2t \ cos(bt^2)[/tex]  and value of b and k are [tex]\pi/2[/tex]  and  [tex]3/2\pi[/tex] respectively.

Learn more about the function here:

brainly.com/question/5245372

Step-by-step explanation:

The coordinates of the feasible region are:(In clockwise direction)

(200, 200)

(300, 200)

(500, 0)

(300, 0)

Answer:

A

Step-by-step explanation:

This shape is a trapezoid. We can divide into two parts: a triangle and a rectangle.

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let A' be the area of the triangle.

● A'= (b*h)/2

b is the base and h is teh heigth.

b= 26-20 = 6 mm

● A'= (6*14)/2 = 42 mm^2

●●●●●●●●●●●●●●●●●●●●●●●●

Let A" be the area of the rectangle.

A"= L*w

L is the length and w is the width.

A"= 14*20

A"= 280 mm^2

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let A be the area of the trapezoid.

A= A'+A"

A= 42+280

A= 322 mm^2

Answer:

Step by step explanation

1. [tex] \frac{1}{1 - x} + \frac{x}{ {x}^{2} - 1} [/tex]

Use [tex] \frac{ - a}{b} = \frac{a}{ - b} = - \frac{a}{b} [/tex] to rewrite the fractions

[tex] - \frac{1}{x - 1} + \frac{x}{(x - 1)(x + 1)} [/tex]

Write all numerators above the Least Common Denominator ( X - 1 ) ( X + 1 )

[tex] \frac{ - (x + 1) + x}{(x - 1)(x + 1)} [/tex]

When there is a ( - ) in front of an expression in parentheses , change the sign of each term in the expression

[tex] \frac{ - x - 1 + x}{(x - 1)(x + 1)} [/tex]

Using [tex](a - b)(a + b) = {a}^{2} - {b}^{2} [/tex] , simplify the product

[tex] \frac{ - x - 1 + x}{ {x}^{2} - 1 } [/tex]

Since two opposites add up to zero, remove them from the expression

[tex] \frac{ - 1}{ {x}^{2} - 1} [/tex]

So, Option A is the right option.

___________________________________

2.

[tex] \frac{ {x}^{2} - x - 12}{ {x}^{2} - 16} - \frac{1 - 2x}{x + 4} [/tex]

Write - X as a difference

[tex] \frac{ {x}^{2} + 3x - 4x - 12 }{ {x}^{2} - 16 } - \frac{1 - 2x}{x + 4} [/tex]

Using [tex] {a}^{2} - {b}^{2} = (a - b)(a + b)[/tex] , factor the expression

[tex] \frac{ {x}^{2} + 3x - 4x - 12 }{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]

Factor the expression

[tex] \frac{x(x + 3) - 4(x + 3)}{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]

Factor out X+3 from the expression

[tex] \frac{(x + 3)(x - 4)}{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]

Reduce the fraction with x-4

[tex] \frac{x + 3}{x + 4} - \frac{1 - 2x}{x + 4} [/tex]

Write all the numerators above the common denominator

[tex] \frac{x + 3 - ( 1- 2x)}{x + 4} [/tex]

When there is a (-) in front of an expression in parentheses, change the sign of each term in the expression

[tex] \frac{x + 3 - 1 + 2x}{x + 4} [/tex]

Collect like terms

[tex] \frac{3x + 3 - 1}{x + 4} [/tex]

Subtract the numbers

[tex] \frac{3x + 2}{x + 4} [/tex]

Undefined at,

X + 4 = 0

Move constant to R.H.S and change its sign

[tex]x = 0 - 4[/tex]

Calculate

[tex]x = - 4[/tex]

So, the answer is :

[tex] \frac{3x + 2}{x + 4} [/tex] , undefined at X = -4 and 4

Hope this helps..

Best regards!!

Answer:

Sine angle of <ACB = 38.68°

Step-by-step explanation:

Hello,

To solve this problem, we need a good representation of the sides and the angle.

See attached document for better illustration.

Assuming it's a right angled triangle,

AC = hypothenus

AB = opposite

BC = adjacent

AC = 40

BC = 25

AB = 25

From trigonometric ratios

Sinθ = opposite/ hypothenus

Sinθ = AB / AC

Sinθ = 25 / 40

Sinθ = 0.625

θ = sin⁻¹0.625

θ = 38.68°

Sine angle of <ACB = 38.68°

Answer:

if its according to the graph but not according to the function then

domain = {x:-4<x<8}

range = {y:y<16}

Answer:

Look at the image below

Answer:

Step-by-step explanation:

cosine law

a²=b²+c²-2(bc)cos44  (c=20, b=15)

a²=15²+20²-2(15*20)cos44

a=√15²+20²-2(15*20)cos44

a=13.91

Answer:

(x - 7)² + (y - 4)² = 49

Step-by-step explanation:

Given

Equation: x² + y² = 49

Required:

New Equation when translated 7 units right and 4 units up

Taking it one step at a time.

When the equation is translated 7 units right, this implies a negative unit along the x axis.

The equation becomes

(x - 7)² + y² = 49

When the equation is translated 4 units up, this implies a negative unit along the y axis.

(x - 7)² + (y - 4)² = 49

The expression can be further simplified but it's best left in the form of

(x - 7)² + (y - 4)² = 49

Answer:

2850

Step-by-step explanation:

We know that only 19% of people prefer the sedans, so we have to find 19% of 15,000. Set up a proportion: [tex]\frac{19}{100}=\frac{x}{15000}[/tex], cross multiply and get 2850.

Answer:

2850 sedans per month

Step-by-step explanation:

19 % want sedans

They expect to sell 15000 cars

Take the number of cars and multiply by the percent sedans

19% * 15000

Change to decimal form

.19*15000

2850

Step-by-step explanation:

[tex]a + b = 6[/tex]

[tex]ab = 1 \times 5 = 5[/tex]

[tex]a = 1 \: \: \: \: \: \: \: \: b = 5[/tex]

[tex]( {x}^{2} + x) + (5x + 5)[/tex]

[tex]x(x + 1) + 5(x + 1)[/tex]

[tex](x + 1)(x + 5)[/tex]

Hope this is correct and helpful

HAVE A GOOD DAY!

Answer:

1

Step-by-step explanation:

There is only one 3 on the cube

Answer:

C = 38n + 1750; 15,050

Step-by-step explanation:

We know that for each person, there's a fee of 38. That signifies that the n will be after 38. 1,750 is a one-time fee, so that's by itself. Plug it into the equation to get your first answer. Now, solve for b) by writing C = 38(350) + 1750; C = 15,050

Answer:

C = 38n + 1750; 15,050

Step-by-step explanation:

brainlist plzzzzzz

Replace f(x) with 0 and solve for x. We do this because the x intercepts always occur when y = 0. Keep in mind that y = f(x).

f(x)=x^3-9x^2+20x

0=x^3-9x^2+20x

x^3-9x^2+20x = 0

x(x^2-9x+20) = 0 .... factor out GCF x

x(x-5)(x-4) = 0 ... factor the stuff inside

x = 0 or x-5 = 0 or x-4 = 0 ... zero product property

x = 0 or x = 5 or x = 4

Step-by-step explanation:

[tex]A[/tex]- all of the chosen balls are white [tex]E_{i}-[/tex] result of the die roll is [tex]i, \quad i \in\{1,2,3,4,5,6\}[/tex]

Probabilities:

since the die is fair:

[tex] P\left(E_{i}\right)=\frac{1}{6} \quad \text { for } \quad i \in\{1,2,3,4,5,6\} [/tex]

If the die rolls [tex]i[/tex] we choose a combination of [tex]i[/tex] balls, among 10 black and five white balls, therefore

\[ \begin{array}{c}

P\left(A \mid E_{1}\right)=\frac{\left(\begin{array}{c} 5 \\

1 \end{array}\right)}{\left(\begin{array}{c} 15 \\ 1 \end{array}\right)}=\frac{5}{15}=\frac{1}{3} \\ P\left(A \mid E_{2}\right)=\frac{\left(\begin{array}{c} 5 \\ 2 \end{array}\right)}{\left(\begin{array}{c} 15 \\ 2

\end{array}\right)}=\frac{10}{105}=\frac{2}{21} \\

P\left(A \mid E_{3}\right)=\frac{\left(\begin{array}{c} 5 \\ 3

\end{array}\right)}{\left(\begin{array}{c} 15 \\ 3

\end{array}\right)}=\frac{10}{455}=\frac{2}{91} \\

P\left(A \mid E_{4}\right)=\frac{\left(\begin{array}{c} 5 \\ 4

\end{array}\right)}{\left(\begin{array}{c} 15 \\ 4

\end{array}\right)}=\frac{1}{273} \\

P\left(A \mid E_{5}\right)=\frac{\left(\begin{array}{c} 5 \\ 5

\end{array}\right)}{\left(\begin{array}{c} 15 \\ 5

\end{array}\right)}=\frac{1}{3003} \\

P\left(A \mid E_{6}\right)=\frac{\left(\begin{array}{c} 5 \\ 6

\end{array}\right)}{\left(\begin{array}{c} 15 \\ 6

\end{array}\right)}=0

\end{array} \]