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tatyana61 [14]
3 years ago
7

Solve T=L(7 + RS) for S. Please help

Mathematics
1 answer:
nikklg [1K]3 years ago
7 0

multiply the ones in bracket by the L

T = 7× L + L × RS

T = 7L + LRS

making S the subject by transferring 7L to the left hand side of the equation.

thus

T - 7L = LRS

dividing through by LR

\frac{T}{LR}  -  \frac{7L}{LR}  = S

S=  \frac{T}{LR}  -  \frac{7}{R}

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katovenus [111]

Answer:

24 square feet

Step-by-step explanation:

Given:

Dorian is spreading mulch on his triangular flowerbed.

The base is 8 feet.

The height is 6 feet.

Question asked:

How many square feet of mulch will he need for the flowerbed?

Solution:

The base of  triangular flowerbed = 8 feet.

The height of  triangular flowerbed = 6 feet.

now, to find how many square feet of mulch he will need for the flowerbed, we will find area of the triangular flowerbed:-

As we know:

Area\ of\ triangle=\frac{1}{2} \times base\times height

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Thus, 24 square feet of mulch he will need for the flowerbed.

8 0
3 years ago
If sine of x equals square root of 2 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences.
kolbaska11 [484]
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tan x = opposite / adjacent..

cos x = square root (2) / 2
tan x = square root (2) / <span>square root (2)  = 1

So, </span>cos x = square root (2) / 2 and tan x<span> = 1</span>
7 0
4 years ago
Which two square root are used to estimate √8?
yan [13]
C is the correct answer
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3 years ago
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ycow [4]

Answer:

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Step-by-step explanation:

plug 7 in for x

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hope this helps!!!

6 0
3 years ago
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How do you use the associative property when you break apart addends
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Associative property of addition is used when you want to group addends. This is mainly used to cluster set of numbers or in this case, addends. How do you use the associative property when you break apart addends? Simple you group them using the open and closed parentheses or brackets. Take for an example 1 + 1 + 2 = 4. Using the associative property you can have either (1 + 1) + 2 = 4 or 1 + (1 + 2) = 4 clustered into place.  



8 0
4 years ago
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