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JulijaS [17]
3 years ago
9

A science class is tracking the progress of plant growth. The class starts the experiment with a plant five centimeters high. Th

e plant grows two centimeters each day. The model for plant growth "y" is given by: y = 2x + 5. What is the meaning of the y-intercept in this equation?
A) the y-intercept is the starting date
B) the y-intercept is two times larger than five
C) the y-intercept is the starting height of the plant
D) the y-intercept is the largest height the plant can grow
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

A

Step-by-step explanation:

when growth of plant is zero means x=0 then y=2(0)+5 = 5

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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

 

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ArbitrLikvidat [17]

Answer:

A. The two lines are neither parallel nor perpendicular.

Step-by-step explanation:

8 0
3 years ago
1. Which of the following number sentences is true?
Anastasy [175]

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2 years ago
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For the real-valued functions f(x)=2x+1 and g(x)=sqrt(x-1), find the composition f o g and specify its domain using interval not
Anettt [7]

Answer:

fog = 2√(x-1) + 1

Domain = [1, \infty)

Step-by-step explanation:

Given the functions  f(x)=2x+1 and g(x)=sqrt(x-1), we are to find the composite function fog

fog = f(g(x))

f(g(x)) = f(√(x-1))

f(√(x+1)) means that we are to replace variable x in f(x) with the function √(x-1)

f(√(x-1)) = 2(√(x-1))+1

f(√(x+1)) = 2√(x-1) + 1

fog = 2√(x-1) + 1

<em>For the function to exist on any real valued function, then the function inside square root i.e x-1 must be greater than or equal to zero (x-1≥0)</em>

If x-1≥0

x≥0+1

x≥1

This means the range of variable x must be values of x greater than or equal to 1.

Domain = [1, \infty)

8 0
3 years ago
Find the slope of the line containing the two points.<br><br> (2,5),(-3,-2)<br><br> Please show work
HACTEHA [7]

Answer:

7/5

Step-by-step explanation:

To solve this problem we will use the formula to find the slope of a line given two points (x₁, y₁), (x₂, y₂)

The formula is:

m= \frac{y{_2} -y_{1}}{x_{2}-x_{1}  } \\\\

So, we're going to use the points (2,5), (-3, -2) in this formula.

m= \frac{-2-(5)}{-3-(2)} \\m=\frac{-2-5}{-3-2}\\m=\frac{-7}{-5} \\m=\frac{7}{5}

Therefore, the slope is 7/5

8 0
3 years ago
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