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Bad White [126]

4 years ago

13

Rodney is helping to make hamburgers for the football game. He's made 10 hamburgers so far. His coach asked him to make at least

30 hamburgers but no more than 80. Solve the inequality and interpret the solution.

1 answer:

anastassius [24]4 years ago

3 0

Answer:

40 I had this question before

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Which of these scales are equivalent to 3cm to 4 km

sergij07 [2.7K]

Answer:

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4 years ago

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andrezito [222]

You need to find the angles and find the sides by using a^2+b^2=c^2

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3 years ago

A print company will print business cards for $.10 each plus a setup charge of $12. Another print company offers business cards

bonufazy [111]

Let the number of cards equal x.

Set up an equation for both companies, set equal to each other and solve for x.

0.10x + 12 = 0.12x + 10

Subtractb0.10x from both sides:

12 = 0.02x + 10

Subtract 10 from both sides:

2 = 0.02x

Divide both sides by 0.02:

X = 2/0.02

X = 100

100 cards would be the same price for both companies.

3 0

3 years ago

I’ll give brainliest help

Alika [10]

Answer:
D (QUESTION AT THE BOTTOM)

Explanation:
C decreases on the Y axis by 3
C increases in the X axis by 6

Doing this on every plot line will move the line ABC to line A’B’C.

Remember:
Y = rise
X = run

6 0

3 years ago

Calculus 2 Master needed, show steps with partial fraction decomposition <img src="https://tex.z-dn.net/?f=%5Cint%283x%5E2-26x%2

Vladimir79 [104]

Answer:

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

Step-by-step explanation:

The fraction will be split into a sum of two other fractions.

The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.

The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.

$\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}$

Combine the two fractions back into one using the common denominator.

$\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}$

This numerator will equal the original numerator.

$A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26$

Match the coefficients.

$A+B=3\\C-5B=-26\\4A-5C=26$

Solve the system of equations.

$A=-1\\B=4\\C=-6$

So we can rewrite the integral as:

$\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx$

Solving:

$\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx$

$-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx$

$-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C$

$-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C$

5 0

3 years ago