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Pie
3 years ago
12

Solve: 2a+2b=c for b. Help me solve this, and thanks if you try.

Mathematics
2 answers:
andrew-mc [135]3 years ago
6 0
2a+2b=c make b the subject 
(-2a) from both sides
2b=c-2a 
Divide by 2 both sides 
b=(c-2a)/2
chubhunter [2.5K]3 years ago
4 0
2a+2b=c \\
2b=c-2a \\
b=\frac{c-2a}{2} = \frac{c}{2} - \frac{2a}{2}=\underline{\underline{\frac{1}{2}c-a}}
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What is the value of x in this diagram? (PLEASE ANSWER!)
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3 years ago
What is the answer to (t+8)(-2)=12
Lunna [17]
(t + 8) (-2) = 12

Distributive property >>>      -2t + -16 = 12

Inverse operation >>>>    -2t + -16+16 = 12+16
 
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8 0
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Read 2 more answers
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Step-by-step explanation:

5 0
3 years ago
NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

8 0
2 years ago
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