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rosijanka [135]
3 years ago
14

What is the slope of 8x - 5y = 20?

Mathematics
2 answers:
Gnom [1K]3 years ago
8 0

Answer: 85

Step-by-step explanation:

harkovskaia [24]3 years ago
7 0

Answer:

8/5 or 1.6

Step-by-step explanation:

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Calculate the average rate of change over the intervals [0, 20] and [60, 80]. What does the average rate on these intervals tell
Assoli18 [71]

\frac{20 - 80}{0 - 60 }  = slope \: or the \: rate \: of \: \\  change \:  \\  \\      \frac{ - 60}{ - 60}  = 1
It tells you that over these intervals the balloon is going at a constant speed.
4 0
3 years ago
An experiment is set up to study the weights of 35 mice after they are injected with a drug. The population mean is 23.50 grams
scZoUnD [109]
Ok you need to add 23.50+3.40
what do you get?
26.9
Round it to the nearest hundreth what do you get?
26.10 is your final answer

7 0
3 years ago
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What is the linear function equation represented by the graph?
Alex Ar [27]
Y=-5/3x+1
I hope this helps!
5 0
3 years ago
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A shoe store recorded the price (in dollars) of 175 pairs of women's shoes and 175 pairs of men's shoes. The data collected was
arsen [322]

Answer:

Most of the men's shoes in the data set cost more than women's shoes.

Step-by-step explanation:

From what I could tell it can't be The price of men's shoes has a higher interquartile range than that of women's shoes. becuse that's false what looks to fit is Most of the men's shoes in the data set cost more than women's shoes.   since on the graph more of the mens  vaules are bigger  

5 0
3 years ago
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
4 0
3 years ago
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