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PtichkaEL [24]
3 years ago
12

Solve the formula for the indicated variable. T(r+s)=rs, for r (Simplify your answer.)

Mathematics
2 answers:
podryga [215]3 years ago
7 0

Hello!

The distributive property is going to be very important in this. It states that when you have a variable multiplied by a set of parenthesis, you distribute that variable inside the parenthesis (as long as the operations inside the parenthesis are addition or subtraction). For example, 2(x + 4), you can distribute the 2 to x and 4, to get 2x + 2(4).

Now, distribute t to inside the parenthesis.

T(r + s) = rs

Tr + Ts = rs

Now, move everything with an r in it to the right side.

Tr + Ts = rs

Ts = rs - Tr

Use the opposite of the distributive property, to extract r from the right side.

Ts = rs - Tr

Ts = r (s - T)

And divide both sides by s - T to isolate the r.

Ts = r (s - T)

\frac{Ts}{s-T} = r

Hope this helps!

yaroslaw [1]3 years ago
5 0

Answer:

R = \frac{-TS}{(T-S)}

Step-by-step explanation:

First, distribute all terms

T(R+S) =RS\\TR + TS =RS

Get all terms with R onto the same side

TR + TS = RS\\TR - RS = -TS

Factor out the R

R(T-S)=-TS

Divide by (T-S) to isolate R

R = \frac{-TS}{(T-S)}

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the number of men is five eights of the number of women working in a factory if there is 24 more women than men how many workers
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Answer:

104

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women = 8/8

men = 5/8

all workers = 8/8 + 5/8 = 13/8

extra women workers = 8/8 - 5/8 = 3/8

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the equation of any line is y = ax + b
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Let A = {a, b, c}, B = {b, c, d}, and C = {b, c, e}. (a) Find A ∪ (B ∩ C), (A ∪ B) ∩ C, and (A ∪ B) ∩ (A ∪ C). (Enter your answe
wariber [46]

Answer:

(a)

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

(b)

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

(c)

(A - B) - C = \{a\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

<em></em>

Step-by-step explanation:

Given

A= \{a,b,c\}

B =\{b,c,d\}

C = \{b,c,e\}

Solving (a):

A\ u\ (B\ n\ C)

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ (A\ u\ C)

A\ u\ (B\ n\ C)

B n C means common elements between B and C;

So:

B\ n\ C = \{b,c,d\}\ n\ \{b,c,e\}

B\ n\ C = \{b,c\}

So:

A\ u\ (B\ n\ C) = \{a,b,c\}\ u\ \{b,c\}

u means union (without repetition)

So:

A\ u\ (B\ n\ C) = \{a,b,c\}

Using the illustrations of u and n, we have:

(A\ u\ B)\ n\ C

(A\ u\ B)\ n\ C = (\{a,b,c\}\ u\ \{b,c,d\})\ n\ C

Solve the bracket

(A\ u\ B)\ n\ C = (\{a,b,c,d\})\ n\ C

Substitute the value of set C

(A\ u\ B)\ n\ C = \{a,b,c,d\}\ n\ \{b,c,e\}

Apply intersection rule

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C)

In above:

A\ u\ B = \{a,b,c,d\}

Solving A u C, we have:

A\ u\ C = \{a,b,c\}\ u\ \{b,c,e\}

Apply union rule

A\ u\ C = \{b,c\}

So:

(A\ u\ B)\ n\ (A\ u\ C) = \{a,b,c,d\}\ n\ \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

<u>The equal sets</u>

We have:

A\ u\ (B\ n\ C) = \{a,b,c\}

(A\ u\ B)\ n\ C = \{b,c\}

(A\ u\ B)\ n\ (A\ u\ C) = \{b,c\}

So, the equal sets are:

(A\ u\ B)\ n\ C and (A\ u\ B)\ n\ (A\ u\ C)

They both equal to \{b,c\}

So:

(A\ u\ B)\ n\ C = (A\ u\ B)\ n\ (A\ u\ C)

Solving (b):

A\ n\ (B\ u\ C)

(A\ n\ B)\ u\ C

(A\ n\ B)\ u\ (A\ n\ C)

So, we have:

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d\}\ u\ \{b,c,e\})

Solve the bracket

A\ n\ (B\ u\ C) = \{a,b,c\}\ n\ (\{b,c,d,e\})

Apply intersection rule

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ \{b,c,e\}

Solve the bracket

(A\ n\ B)\ u\ C = \{b,c\}\ u\ \{b,c,e\}

Apply union rule

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = (\{a,b,c\}\ n\ \{b,c,d\})\ u\ (\{a,b,c\}\ n\ \{b,c,e\})

Solve each bracket

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}\ u\ \{b,c\}

Apply union rule

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

<u>The equal set</u>

We have:

A\ n\ (B\ u\ C) = \{b,c\}

(A\ n\ B)\ u\ C = \{b,c,e\}

(A\ n\ B)\ u\ (A\ n\ C) = \{b,c\}

So, the equal sets are:

A\ n\ (B\ u\ C) and (A\ n\ B)\ u\ (A\ n\ C)

They both equal to \{b,c\}

So:

A\ n\ (B\ u\ C) = (A\ n\ B)\ u\ (A\ n\ C)

Solving (c):

(A - B) - C

A - (B - C)

This illustrates difference.

A - B returns the elements in A and not B

Using that illustration, we have:

(A - B) - C = (\{a,b,c\} - \{b,c,d\}) - \{b,c,e\}

Solve the bracket

(A - B) - C = \{a\} - \{b,c,e\}

(A - B) - C = \{a\}

Similarly:

A - (B - C) = \{a,b,c\} - (\{b,c,d\} - \{b,c,e\})

A - (B - C) = \{a,b,c\} - \{d\}

A - (B - C) = \{a,b,c\}

<em>They are not equal</em>

4 0
3 years ago
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