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2 years ago

What is the root of the quadratic equation y= (x-4) (x+7)

Mathematics

1 answer:

pishuonlain [190]2 years ago

4 0

Root are also known as what x is equal to. So, to find x you would need to set (x-4)=0 and (x+7)=0
X=4. X=-7

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Give the slope and the y-intercept of the line y = – 6x– 4. Make sure the y-intercept is written as

jonny [76]

Answer:

Step-by-step explanation:

Slope = -6

Y-intercept (0, -4)

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3 years ago

HELPPPP will mark brainliest

ArbitrLikvidat [17]

Answer:

12x-2

Step-by-step explanation:

You need to add 5x+1 and 7x-3.

5x+1+7x-3.

We need to organize this into common terms.

5x+7x+1-3

Now we simplify.

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3 years ago

A circle has a sector with area 33 pi and a central angle of 11/6 pi radians. What is the area of a circle?

Mashutka [201]

Answer:

36π

Step-by-step explanation:

The area of a circle is given as:

$A = \pi r^2$

where r = radius of the circle

The area of a sector of a circle is given as:

$A_s = \frac{\alpha }{2\pi} * \pi r^2$

where α = central angle in radians

Since $\pi r^2$ is the area of a circle, A, this implies that:

$A_s = \frac{\alpha }{360} * A$

A circle has a sector with area 33 pi and a central angle of 11/6 pi radians.

Therefore, the area of the circle, A, is:

$33 \pi = \frac{\frac{11 \pi}{6} }{2 \pi} * A\\\\33\pi = \frac{11}{12} * A\\\\=> A = \frac{33\pi * 12}{11}\\ \\A = \frac{396 \pi}{11} \\\\A = 36\pi$

The area of the circle is 36π.

5 0

3 years ago

Read 2 more answers

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

$V = \displaystyle\frac{1}{3}\pi r^2 h$

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)$

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$