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3 years ago

1/2 (4a + 10b) = c solve for b Can someone help me on this

Mathematics

2 answers:

Butoxors [25]3 years ago

7 0

Here is the answer I got: B=c/5 - 2a/5

Artyom0805 [142]3 years ago

7 0

1/2 (4a + 10b) = c

Remove the 1/2 by multiplying both sides by 2:

4a + 10b = 2c

Subtract 4a from both sides:

10b = 2c -4a

Divide both sides by 10:

b = (2c-4a)/10

SImplify by the fraction:

b = (c - 2a)/5

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Solve the compound inequality. 3x + 1 <4 or 9x > 31

WINSTONCH [101]

Answer:

Step-by-step explanation:

3x < 4 - 1 or x > 31/9

3x < 3 or x > 31/9

x < 1 or x > 31/9

7 0

3 years ago

Ticket for football final selling well, on Thursday 4/7 of the tickets are sold, on Friday 1/4 of the tickets are sold, What fra

sergij07 [2.7K]

Answer:

5/28

Step-by-step explanation:

1/1 represents the whole amount of tickets.

4/7 are sold on Thursday.

an additional 1/4 are did on Friday.

fraction of available tickets on Saturday are therefore

1/1 - 4/7 - 1/4 = 4/1 - 16/7 - 1/1 = 28/1 - 16/1 - 7/1 = 5/1

that means out of 28 equal parts of the total number of tickets 16+7 = 23 parts have been sold.

so, there are only 5 out of the total 28 parts available on Saturday, and the fraction of available tickets is therefore 5/28.

we could have also tried to bring both fractions to the same denominator, and found that this is 28.

and then

4/7 + 1/4 = 4/4 × 4/7 + 7/7 × 1/4 =

= 16/28 + 7/28 = 23/28

and the whole is 28/28.

so we have 28/27 - 23/28 = 5/28 left.

surprise, surprise, it is the same result ...

4 0

2 years ago

Solve for x. 3(2x+2)+x+5=-10

Paul [167]

3(2x+2)+x+5=-10
6x+6+x+5=-10
7x+11=-10
7x=-21
x=-3

8 0

3 years ago

Read 2 more answers

he volume of a fish tank is 50 cubic feet. If the density is 0.2 fish over feet cubed, how many fish are in the tank?

V125BC [204]

There are 10 fish in the tank

Solution:

Given that, volume of a tank is 50 cubic feet

Density is 0.2 fish over feet cubed

To find: Number of fish in the tank

Number of fish in the tank is found by multiplying the volume of tank and density

The formula is given by:

$\text{Number of fish} = \tex{\text{volume of tank}} \times {\text{density of fish over feet cubed}}$

Substituting the given values, we get

$\text{Number of fish} = 50 ft^3 \times \frac{\text{ 0.2 fish}}{ft^3}\\\\ \text{Number of fish} = 50 \times 0.2 \text{ fish }\\\\ \text{Number of fish} = 10 \text{ fish }$

Thus there are 10 fish in the tank

5 0

4 years ago

How would I add the fractions? <img src="https://tex.z-dn.net/?f=-2%2F5%20%2B%202%2F3p%20%2B1%2F3p%20%3D%202p%20-%2016%2

LiRa [457]

$\bf -\cfrac{2}{5}+\cfrac{2}{3}p+\cfrac{1}{3}p=2p-\cfrac{16}{5}\implies \stackrel{\textit{LCD of 15}}{-\cfrac{2}{5}+\cfrac{2p}{3}+\cfrac{p}{3}}=\stackrel{\textit{LCD of 5}}{2p-\cfrac{16}{5}}
\\\\\\
\cfrac{(3\cdot -2)+(5\cdot 2p)+(5\cdot p)}{15}=\cfrac{(5\cdot 2p)-16}{5}
\\\\\\
\cfrac{-6+10p+5p}{15}=\cfrac{10p-16}{5}\impliedby \textit{now we cross-multiply}
\\\\\\
-30+50p+25p=150p-240\implies -30+240=150p-75p
\\\\\\
210=75p\implies \cfrac{210}{75}=p\implies \cfrac{14}{5}=p$

8 0

3 years ago