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MrRissso [65]
3 years ago
14

Which description is correct for the x-intercept(s) of Function A and Function B?

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

Step-by-step explanation:

Generally,

A straight line equation is given as

y = mx + c

Where,

m is the slope or gradient of the line

c is the intercept of y axis, when x = 0

Now,

Given the function A

g(x) = 0 7 0 2 -2

x = -2 3. 5 -1 0

So, we said the intercept c is when x = 0,

Therefore, in this case when x=0, g(x) = -2

Then, the intercept is -2 on y axis

Then, it has only one intercept, which is -2 on y-axis

To get the intercept on x axis, we will set y to zero,

In this case,

g(x) = 0, then, we have the intercept on x-axis to be -2 and 5

So it has 2-intercept on x axis, one positive and one negative

Given the function B

F(x) = |x| - 3

Comparing this to equation of a line, y= mx + c,

Then we notice that, the intercept is -3

Then, it has only one intercept which is -3 on y axis.

Now, to get the intercept on x axis, we will set f(x) = 0

F(x) = |x| - 3

0 = |x| - 3

Then, |x| = 3

Then, x = 3 or -x =3

Note that, |x| means x and -x

Then, x = 3 or x = -3

So it has also two intercept on the x axis, one positive and one negative.

Then, the third option is the correct option

Both functions have one negative and one positive x-intercept.

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Choose the system of inequalities that best matches the graph below.
lapo4ka [179]

Answer:

D) y\:, y\:>\:-x

Step-by-step explanation:

The blue boundary line has a y-intercept of 2 and a slope of  2.

It has equation : y=2x+2

Since the boundary line is not solid and the lower half plane of this line is shaded, its inequality is

y\:

The blue boundary line passes through the origin and has slope -1.

Its equation is y=-x.

Since the boundary line is not solid and the right half-plane is shaded, its inequality is y\:>\:-x

The correct choice is D.

4 0
4 years ago
Which postulate proves the two
adoni [48]

Answer:

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7 0
3 years ago
HELP!!<br>What is the surface area of this shape?​
Hunter-Best [27]

2 triangular faces

1/2bh

10x15=150

2x150= 300

‘back’ face

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6 0
3 years ago
A sphere of radius r is cut by a plane h units above the equator, where
Anika [276]
Consider the top half of a sphere centered at the origin with radius r, which can be described by the equation

z=\sqrt{r^2-x^2-y^2}

and consider a plane

z=h

with 0. Call the region between the two surfaces R. The volume of R is given by the triple integral

\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx

Converting to polar coordinates will help make this computation easier. Set

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

Now, the volume can be computed with the integral

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta

You should get

\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)
5 0
3 years ago
Find the range of {(3.2, –3), (7.6, 5.9), (1.4, –3), (–9.1, 8.3)}.
MrRa [10]

Answer:

{-3, 5.9, -3, 8.3}

Step-by-step explanation:

The range is always the second number of all pairs

8 0
3 years ago
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