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3 years ago

Help fast please!!!!!!!!

Mathematics

1 answer:

MArishka [77]3 years ago

4 0

Step-by-step explanation:

The area is

${x}^{2} - 110x + 2800 \\ = {x}^{2} - 40x - 70x + 2800 \\ = x(x - 40) - 70(x - 40) \\ = (x - 40)(x - 70)$

Since the width is

$x - 40 \: (feet)$

Then, the length will be

$x - 70 \: (feet)$

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What’s 18 divided by 7?

GalinKa [24]

Here is the answer...

2.57142857143

7 0

4 years ago

What is an equation of the line that passes through the points ( 0,-7) and ( 5, -1)

tresset_1 [31]

Answer:

y = 5/6x - 5.167

y = 5/6x - 31/6

Step-by-step explanation:

The equation of a line

y = mx + c

Step 1: find the slope

m = y2 - y1 / x2 - x1

Given points

( 0 , -7) ( 5 , -1)

x1 = 0

y1 = -7

x2 = 5

y2 = -1

Insert the values into the equation

m = -1 - (-7) / 5 - 0

m = -1 + 7 / 5

m = 6/5

y = 6/5x + c

Step 2: sub any of the two points given into the equation

y = 5/6x + c

( 5 , -1)

x = 5

y = -1

-1 = 5/6(5) + c

-1 = 5*5/6 + c

-1 = 25/6 + c.

c = -1 - 25/6

LCM = 6

c = -6 - 25 / 6

c = -31/6

c = -5.167

Step 3: sub c into the equation

y = 5/6x + c

y = 5/6x - 31/6

y = 5/6x - 5.167

The equation of the line is

y = 5/6x - 5.167

7 0

3 years ago

Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the

Rama09 [41]

Question:

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s 2s 3s 4s

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

$h(t) = -16t^2 + v_0t+h_0$

Initial velocity = 27 feet/sec

$h_0 = 10\ feet$

Therefore,

$h(t) = -16t^2 +27t+10$

At the point the object hits the ground, h(t) = 0

$-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0$

Solve by quadratic formula,

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125$