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Sunny_sXe [5.5K]

3 years ago

13

The given graph represents the function f(x) = 2(5)x. How will the appearance of the graph change if the a value in the function

is decreased, but remains greater than 0?

2 answers:

Naddika [18.5K]3 years ago

6 0

Answer:

answer is C

Step-by-step explanation:

castortr0y [4]3 years ago

4 0

The answer is C: "The graph will show an initial value that is lower on the y-axis."

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3 years ago

250% of what number 152.5

Nadusha1986 [10]

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3 years ago

Evaluate g(x) = 4 – 3x when x<br> = -3, 0, and 5.

nasty-shy [4]

G(x) when x = -3 : 4 - 3(-3) = 13
2. 4
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5 0

3 years ago

Write the equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x

Solnce55 [7]

Answer:

The equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x is $\mathbf{y=-\frac{2}{3}x+3 }$

Step-by-step explanation:

We need to Write the equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x

The equation in slope-intercept form is: $y=mx+b$ where m is slope and b is y-intercept.

Finding Slope:

The both equations given are parallel. So, they have same slope.

Slope of given equation y= - 2/3x is m = -2/3

This equation is in slope-intercept form, comparing with general equation $y=mx+b$ where m is slope , we get the value of m= -2/3

So, slope of required line is: m = -2/3

Finding y-intercept

Using slope m = -2/3 and point (-3,5) we can find y-intercept

$y=mx+b\\5=-\frac{2}{3}(-3)+b\\5=2+b\\ b=5-2\\b=3$

So, we get b = 3

Now, the equation of required line:

having slope m = -2/3 and y-intercept b =3

$y=mx+b\\y=-\frac{2}{3}x+3$

The equation of the line in slope intercept form that passes through the point (-3, 5) and is parallel to y= - 2/3x is $\mathbf{y=-\frac{2}{3}x+3 }$

5 0

3 years ago

Read 2 more answers

A ball is thrown in the air from a height of 3 m with an initial velocity of 12 m/s. To the nearest tenth of a second, how long

stepladder [879]

Answer: 1.4 seconds

<u>Step-by-step explanation:</u>

The equation is: h(t) = at² + v₀t + h₀ where

a is the acceleration (in this case it is gravity)
v₀ is the initial velocity
h₀ is the initial height

Given:

a = -9.81 (if it wasn't given in your textbook, you can look it up)
v₀ = 12
h₀ = 3

Since we are trying to find out when it lands on the ground, h(t) = 0

EQUATION: 0 = 9.81t² + 12t + 3

Use the quadratic equation to find the x-intercepts

a=-9.81, b=12, c=3

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(12)\pm \sqrt{(12)^2-4(-9.81)(3)}}{2(-9.81)}\\\\\\x=\dfrac{-12\pm 16.2}{-19.62}\\\\\\x=\dfrac{-12+ 16.2}{-19.62}=-0.2\qquad x=\dfrac{-12- 16.2}{-19.62}=\large\boxed{1.4}\\$

Note: Negative time (-0.2) is not valid

4 0

3 years ago