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barxatty [35]
3 years ago
6

Diego and Jada are both trying to write an expression with fewer terms that is equivalent to

Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer: Diego, 4a + 9b is equivalent

Step-by-step explanation:

7a + 5b - 3a + 4b

= (7a - 3a) + (5b + 4b)           We rerange the variables, and put them together with (), which make it more clear to solve.

= 4a + 9b

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Select the correct answer.<br> What are the roots of 12x = 9 + 5x2?
Lorico [155]

Answer:

12x=9+5x^{2}

12x=9+5x2

Step-by-step explanation:

5 0
3 years ago
The container that holds the water for the football team is
Rom4ik [11]

Answer:

30 Gallons

Step-by-step explanation:

Okay so you need to start by making an equation for yourself. Since the container starts at 1/10 of the total, you want to start by saying 1/10x. Then you add 12 gallons of water to make it half full. So 1/10x + 12 = 1/2x where x is the total amount that can fit in the container.

1/10x + 12 = 1/2x (Move your x's to the same side.)

12 = 1/2x - 1/10x (Find a common denominator.)

12 = 5/10x - 1/10x (Simplify/)

12 = 4/10x (Multiply both sides by 10; your demoninator)

120 = 4x (Divide)

x = 30

Your container holds 30 gallons.

3 0
3 years ago
HELP ME PLEASEEEEEE ASAPPPPPP
krok68 [10]

your answer is 10 1/8.

I hope this helps

4 0
3 years ago
I have no idea who to do this?
natka813 [3]
Click the "ask for help" button
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3 0
3 years ago
This question is designed to be answered without a calculator. Use the antiderivative formula shown, where f(u) represents a fun
Phoenix [80]

Answer:

f(u) = u

Step-by-step explanation:

Given

\int {\ln(u)} \, du = u[\ln(u) - f(u)] + c

Required

Find f(u)

To do this, we start by integrating the left-hand side

\int {\ln(u)} \, du = u\ln(u) - f(u)+ c

Using integration by parts, we have:

\int {fg'}  = fg - \int f'g

So, we have:

f = \ln(u)

Differentiate

f'=\frac{1}{u}

g' = 1

Integrate

g=u

So:

\int {fg'}  = fg - \int f'g

\int \ln(u)\ du = \ln(u) * u - \int \frac{1}{u} * u\ du

\int \ln(u)\ du = u\ln(u)  - \int \ du

So, we have:

u\ln(u)  - \int \ du = u\ln(u) - f(u) + c

Integrate du using constant rule

u\ln(u)  - u + c = u\ln(u) - f(u) + c

Subtract c from both sides

u\ln(u)  - u = u\ln(u) - f(u)

Subtract u ln(u) from both sides

- u =  - f(u)

Rewrite as:

f(u) = u

8 0
4 years ago
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