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3 years ago

Need Help. Please! Will Give Brainliest!

Mathematics

2 answers:

Sindrei [870]3 years ago

4 0

The answer is 45.5 please give Brainliest!

Ray Of Light [21]3 years ago

4 0

For the Sean one: 90$

For the Carol one: 46 miles.

I hope this helps you, and please don't forget to say thanks and give brainliest! It really helps out!

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Which ordered pair is a solution of the system ?

drek231 [11]

Hello!

Since we already see an equation that equals y (-2x-4), we can plug that equation into the first to solve for x.

x+4(-2x-4)=19
x-8x-16=19
-7x=35
x=-5

Now that we know the value of x, we will plug it into the second equation to solve for y.

y=-2(-5)-4
y=10=4
y=6

This gives us the answer below

$\left \{ {{x=-5} \atop {y=6}} \right.$

The correct answer is A) (-5,6)

I hope this helps!

7 0

3 years ago

Item 3

NikAS [45]

Answer:

I'm taking the test right now so i'll tell you the right answer after I submit my test but for now I think it's C.

5 0

2 years ago

Is 85/95 greater than 64/76

madam [21]

85/95 is equal to 0.8947
64/76 is equal to 0.842
Therefore, 85/95 is greater

5 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

Help?????????????????

lesantik [10]

Y = -4x
**************************************************************************

8 0

3 years ago

Read 2 more answers