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2 years ago

Find the the value f y in the parallelogram. A.2 B. 1.5 C. 0.5 D. 7.5

Mathematics

1 answer:

g100num [7]2 years ago

7 0

Answer:

Step-by-step explanation:

All parallelograms have diagonals which bisect each other. This means the diagonals intersect and create equal lengths.

5y = 3x + 6

8x - 2 = 4x

Solve for x using 8x - 2 = 4x. Then substitute x into the other equation.

8x - 2 = 4x

-2 = -4x

1/2 = x

Substitute x = 1/2.

5y = 3(1/2) + 6

5y = 3/2 + 6

5y = 3/2 + 12/2

5y = 15/2

y = 15/10

y = 3/2 or 1.5

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At what value of x does graph of the following function f(x) have a vertical asymptote? F(x)= 1/x+3

kirill115 [55]

Answer:

Step-by-step explanation:

Given

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2 years ago

When looking at the graph of a 5th degree function, how can you determine if all of the zeros of the function are real?

Finger [1]

Answer:

If it cuts x-axis 5 times.

Step-by-step explanation:

When we look at the graph of a function we can see its real roots by looking at its graph

The intersecting points that is the number of times a line cutting x-axis will be the real root of the function

So, by looking at the 5th degree function the number of time that function cuts x-axis will be the number of real roots.

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3 years ago

A right circular cylinder has a height of 5in. And a base area of 20 in2. What is the volume of the cylinder

koban [17]

Answer:

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Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

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

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0

3 years ago