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3 years ago

6

I NEED HELP WITH THIS PLEASE!

1 answer:

Sever21 [200]3 years ago

8 0

To find the answer to A, plug in -2 for x. So your equation will be f(-2)=8-4(-2)

Your answer will be 16, I think.

For B, plug in 8 for f(x). The equation should be 8=8-4(x)

The answer is x=0, I think.

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SOMEONE PLEASE HELP ME!!!!!!!

OleMash [197]

Answer:

Step-by-step explanation:

$\frac{57}{40.3}=\frac{23}{QR} \\$

cross multiply

57(QR)=926.9

QR=16.26...

round to nearest tenth

QR=16.3

4 0

3 years ago

Y=2x-1<br><br> y=-3x+14<br><br> Solve for y

LUCKY_DIMON [66]

Y=2x-1
y=-3x+14

First you would substitue one of the y's for the other so
2x-1=-3x+14

no you would solve for x...
add one on both sides
2x=-3x+15

now add 3x to both sides which becomes
5x=15

divide 5 on both sides which gives you x=3

now to solve for y plug into any one of the equations(doesn't matter which one) from before 3 for x and solve
y=2(3)-1
y=6-1
y=5

And that is you answer y=5

3 0

3 years ago

What is 8r + 12p - 7 - 3p and how to get it

MrRa [10]

8r+(12-3)p-7------>8r+9p-7

3 0

3 years ago

The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width

blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given perimeter

perimeter of rectangle = 2 ( length + width )

=> 30.8 km = 2 ( x + y )

=> $x + y = \frac{30.8}{2}$

=> y = 15.4 – x ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,

=> $length^2 + width^2 = diagonal^2$

=> $x^2 + y^2 = 11^2$

=> $x^2 + y^2 = 121$

On substituting value of y from (1) in above equation we get

=> $x^2 + (15.4-x)^2 = 121$

=> $x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121$

=> $2x^2-30.8x + 237.16 -121 = 0$

=> $2x^2-30.8x + 116.16 = 0$

Solving above quadratic equation using quadratic formula

General form of quadratic equation is

$ax^2 +bx +c = 0$

And quadratic formula for getting roots of quadratic equation is

$x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}$

As equation is $2x^2-30.8x + 116.16 = 0$ , in our case

a = 2 , b = -30.8 and c = 116.16

Calculating roots of the equation we get

$x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}$

$x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{(30.8\pm29.33)}4$

$x=\frac{(30.8+29.33)}{4}$

$x=\frac{(30.8-29.33)}{4}$

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0

3 years ago

Mplete the solution of the equation. Find the

Alex777 [14]

Answer:

-2

Step-by-step explanation:

If x=0

-3(0)+2y=-4

-0+2y=-4

We can say

2y-0=-4

2y=-4

Dividing both side by 2,

y=-2

5 0

2 years ago