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geniusboy [140]
3 years ago
8

2(3x + 4) = 4x + 22 what is x

Mathematics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

x = 7

I need brainliest so I can rank up

Step-by-step explanation:

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What is the volume of a right circular cylinder with a diameter of 46.25 cm and a height of 18.5 cm?
alexdok [17]
Radius = <span> <span> <span> 23.125 </span> </span> </span> cm
Cylinder Volume =  <span>π <span>• r² • height
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Cylinder Volume =  3.14 * 23.125^2 * 18.5

Cylinder Volume =  <span> <span> <span> 31,064.53515625 </span> </span> </span>

Cylinder Volume =  <span> <span> <span> 31,064.54 cubic centimeters


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7 0
3 years ago
I need an Answer Immediately pls!!!!!!!!!
Salsk061 [2.6K]

Answer:

2,033-993.50= 1039.50/74.25= 14

Step-by-step explanation:

not sure the exact equation but p would be 14 and you'd do something similar to that shown above

8 0
3 years ago
Two-step equations with decimals and fractions<br><br> Solve for p.<br><br> −5(p+ 3/5 )=−4
GrogVix [38]

Answer:

1/5 or 0.2

Step-by-step explanation:

-5(p+3/5)=-4

Let's convert 3/5 to decimal form:

-5(p+0.6)=-4

Expand parentheses:

-5p-3=-4

Add -3 to both sides:

-5p=-1

Divide both sides by -5:

p=1/5 or 0.2

Hope this helps!

4 0
3 years ago
Read 2 more answers
Suppose <img src="https://tex.z-dn.net/?f=m" id="TexFormula1" title="m" alt="m" align="absmiddle" class="latex-formula"> men and
ollegr [7]

Firstly, we'll fix the postions where the n women will be. We have n! forms to do that. So, we'll obtain a row like:

\underbrace{\underline{~~~}}_{x_2}W_2 \underbrace{\underline{~~~}}_{x_3}W_3 \underbrace{\underline{~~~}}_{x_4}... \underbrace{\underline{~~~}}_{x_n}W_n \underbrace{\underline{~~~}}_{x_{n+1}}

The n+1 spaces represented by the underline positions will receive the men of the row. Then,

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m~~~(i)

Since there is no women sitting together, we must write that x_2,x_3,...,x_{n-1},x_n\ge1. It guarantees that there is at least one man between two consecutive women. We'll do some substitutions:

\begin{cases}x_2=x_2'+1\\x_3=x_3'+1\\...\\x_{n-1}=x_{n-1}'+1\\x_n=x_n'+1\end{cases}

The equation (i) can be rewritten as:

x_1+x_2+x_3+...+x_{n-1}+x_n+x_{n+1}=m\\\\&#10;x_1+(x_2'+1)+(x_3'+1)+...+(x_{n-1}'+1)+x_n+x_{n+1}=m\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-(n-1)\\\\&#10;x_1+x_2'+x_3'+...+x_{n-1}'+x_n+x_{n+1}=m-n+1~~~(ii)

We obtained a linear problem of non-negative integer solutions in (ii). The number of solutions to this type of problem are known: \dfrac{[(n)+(m-n+1)]!}{(n)!(m-n+1)!}=\dfrac{(m+1)!}{n!(m-n+1)!}

[I can write the proof if you want]

Now, we just have to calculate the number of forms to permute the men that are dispposed in the row: m!

Multiplying all results:

n!\times\dfrac{(m+1)!}{n!(m-n+1)!}\times m!\\\\&#10;\boxed{\boxed{\dfrac{m!(m+1)!}{(m-n+1)!}}}

4 0
3 years ago
How do you solve this? cosθ-tanθcosθ=0
laila [671]
First, lets note that tan(\theta)\cdot cos(\theta)=sin(\theta). This leads us with the following problem:

cos(\theta)-sin(\theta)=0

Lets add sin on both sides, and we get:

cos(\theta)=sin(\theta)

Now if we divide with sin on both sides we get:

\frac{cos(\theta)}{sin(\theta)}=1

Now we can remember how cot is defined, it is (cos/sin). So we have:

cot(\theta)=1

Now take the inverse of cot and we get:
\theta=cot^{-1}(1)=\pi\cdot n+ \frac{\pi}{4} , \quad n\in \mathbb{Z}

In general we have cot^{-1}(1)=\frac{\pi}{4}, the reason we have to add pi times n, is because it is a function that has multiple answers, see the picture:

4 0
3 years ago
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