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3 years ago

H(y)=4|y+2|−3, evaluate h(−8).

Mathematics

1 answer:

nikdorinn [45]3 years ago

7 0

Answer:

h(-8)=21

Step-by-step explanation:

-8+2=-6

abs(-6)=6

4*6=24

24-3

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Write the equation of a line that is perpendicular to y=-0.3x+6 and that passes through the point (3,-8)

MariettaO [177]

Hey there!

First, let's look at what perpendicular means. Imagine a cross, where there's all 90 degree angles. That's exactly what we're talking about when we say perpendicular. The given equation is in slope-intercept form, where we have:

y = mx + b
where m is the slope and b is the y-intercept.

When we're writing an equation with a perpendicular slope, we use the negative reciprocal of the given slope. Thus, we can make 0.3 1/3, and take the reciprocal to make 3, and make it negative 3 as it's the negative reciprocal. Now, we know we have a line with the slope of -3 and goes through (-3, 8). We can use the x and y values in this set of points, along with the slope, to create an equation to solve for b. That gives us:

8 = -3(-3) + b
8 = -9 + b
17 = b

Now, since we have slope and y-intercept, we can write our equation as:

y = -3x + 17

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3 years ago

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IS ANYONE GOOD AT CALCULUS? PLEASE HELP!!! The rate at which a certain radioactive isotope decays is given by = negative 0.0032

victus00 [196]

Answer:

90 days.

Step-by-step explanation:

dy/dt = - 0.0032y

dy = -0.0032y dt

dy/y = -0.0032 dt

Integrating:

ln y = -0.0032t + c where c is some constant

y = e^(-0032t) + e^c

y = A e^(-0032t) where A is a constant.

At time 0 we can let y = 1 so in that case A = 1

When there is 0.75 of y left:

0.75 = e^-0032t

ln 0,75 = -0032 t

t = ln 0.75 / -0032

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Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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kvv77 [185]

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Range: $-1 \le y \le 1$ where y is a real number

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The range is the set of y values between -1 and 1 because this is the set of y outputs. The largest y can get is +1 or just 1. The smallest y can get is -1. The value of y can be any number in between as there are no gaps. <span />

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