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2 years ago

11

Please Help! I need help with #16, 18, and 20. You can ignore everything else. I will give Brainliest!

1 answer:

laiz [17]2 years ago

6 0

It’s 15 for #16 and then for #18 it would be 55 then for 20 it would be 6

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Solve for x: 4 over 5 x + 4 over 3 = 2x x = __________________ Write your answer as a fraction in simplest form. Use the "/" sym

alexgriva [62]

Let's see.

4x/5 + 4/3 = 2x

First we have to make each denominator the same, so I'll multiply 4x/5 by 3/3, 4/3 by 5/5, and 2x by 15/15

Now we have 12x/15 + 20/15 = 30x/15

With everything in the same denominator we can solve the new equation of
12x + 20 = 30x

20 = 18x

10 = 9x

X= 10/9

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3 years ago

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Let a, b, and c be constants, and let x be a variable. Which of the following is

Nonamiya [84]

Answer:

C

Step-by-step explanation:

We have:

$a(x+b)$

Where a<0.

We would like to solve for a. So, let’s divide both sides by a.

Notice that a<0. Therefore, a is negative. Hence, we must flip the sign since we are dividing by a negative. This yields:

$x+b>\frac{c}{a}$

Now, we will subtract b from both sides. Therefore, our inequality is:

$x>\frac{c}{a}-b$

Hence, our answer is C.

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2 years ago

A fish aquarium that needs water measures 5ft long by 3 feet wide by 2 feet tall. If Josh only has has 12ft of water, how much m

Pepsi [2]

Answer:

Step-by-step explanation:

18 ft

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3 years ago

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Ms. Parker baked 6 dozen cookies. She gave 2 dozen cookies to her sister, 1 À dozen

Lyrx [107]

Answer:

Step-by-step explanation:

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3 years ago

A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

5 0

3 years ago