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antiseptic1488 [7]
3 years ago
8

Graph the line whose y -intercept is 5 and whose x -intercept is −7 .

Mathematics
1 answer:
Tcecarenko [31]3 years ago
8 0
Look at the vertical line on the graph and find five
Place a dot there
Look at the horizontal line and look to the left to find -7
Place a dot there
Now, just get a ruler and draw a line connecting the two dots
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Which value is the solution of the equation 5y-4x=-3
Lelechka [254]
I hope this helps you

7 0
3 years ago
Fifteen minus three to the power of two
Stells [14]
Your first answer is correct but I’m not sure if your second is. But I’m leaning towards it is not correct.

Hope this helps!

(Don’t forget to mark branliest if it is correct :) )
5 0
2 years ago
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The variables a, b, and c represent polynomials where a = x 1, b = x2 2x − 1, and c = 2x. what is ab c in simplest form?
inna [77]
If you would like to write a * b + c in simplest form, you can do this using the following steps:

a = x + 1
b = x^2 + 2x - 1
c = 2x
a * b + c = (x + 1) * (x^2 + 2x - 1) + 2x = x^3 + 2x^2 - x + x^2 + 2x - 1 + 2x = x^3 + 3x^2 + 3x - 1

The correct result would be x^3 + 3x^2 + 3x - 1.




7 0
3 years ago
Read 2 more answers
PLEASE!! For 22 points!!
daser333 [38]

Answer:

y = 3/5x + 3/5.

Step-by-step explanation:

the /'s are division symbols.              3            3

                                                  y =     _   +       _

                                                           5            5

5 0
2 years ago
Newton's law of cooling is:
Mnenie [13.5K]

Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

7 0
3 years ago
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