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andrezito [222]
2 years ago
5

Adam is comparing the graphs of y=4x and y=8x. Which of the following statements is TRUE?

Mathematics
1 answer:
Inga [223]2 years ago
5 0
The first one is true your welcome hope thi helps
You might be interested in
Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane
postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²            ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i.  Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}

d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}

d = \sqrt{(0)^2+ (0)^2 + (-5)^2}

d = \sqrt{(25)}

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²  

(x - 3)² + (y + 4)² + (z - 5)² = 25  

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25  

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i.  Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}

d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}

d = \sqrt{(-3)^2 + (0)^2+ (0)^2}

d = \sqrt{(9)}

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²  

(x - 3)² + (y + 4)² + (z - 5)² = 9  

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9  

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i.  Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}

d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}

d = \sqrt{(0)^2 + (4)^2+ (0)^2}

d = \sqrt{(16)}

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²  

(x - 3)² + (y + 4)² + (z - 5)² = 16  

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

 

3 0
3 years ago
Need answer urgently please will give thanks and mark Brainliest
babymother [125]

Answer:

D

Step-by-step explanation:

sqrt(7) = about 2.6

and that divided by 2 is

0.8  which means D is the Answer

8 0
2 years ago
Sandra throws an object straight up into the air with an initial velocity of 38 ft/s from a platform that is 30 ft above the gro
Makovka662 [10]

Answer:

The answer is C.

Step-by-step explanation:

Given formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height.

In this case, the initial postion is a platform 30ft above ground so h0=+30

The initial velocity is 38 ft/s straight up into the air so v0=+38

h(t)=-16t2+38t+30

When the object hits the ground, h=0.

h=-16t2+38t+30=0

Simplifying 8t2-19t-15=0

(8t+5)(t-3)=0

t=-5/8 or 3

As time cannot be -ve, t=3s. The answer is C.

6 0
2 years ago
Read 2 more answers
4x−4<8 AND 9x+5>23 Solve for X.
siniylev [52]

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>☝</em><em>✌</em><em>✌</em><em>✌</em><em>✌</em>

5 0
3 years ago
PLZ HELP!!!!!!!!!!!!!
Rasek [7]

Answer:

Move the decimal point 3 times to the left.

Step-by-step explanation:

6 0
2 years ago
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