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2 years ago

Hey can anyone help pls and thank you

Mathematics

2 answers:

mixer [17]2 years ago

6 0

Answer:

x = 16

Step-by-step explanation:

(7x - 4) and (5x + 28) are alternate exterior angles and are congruent, so

7x - 4 = 5x + 28 ( subtract 5x from both sides )

2x - 4 = 28 ( add 4 to both sides )

2x = 32 ( divide both sides by 2 )

x = 16

MrRissso [65]2 years ago

4 0

Answer:

5x+28=7x-4

5x+32=7x

32=2x

x=16

Hope This Helps!!!

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Jeff gave 1/4 of a sum of money to his wife. Then he divided the remainder equally among his 4 children.

Semenov [28]

Answer:

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200

Step-by-step explanation:

Given as :

Let The sum of money did Jeff have = $ x

The fraction of money did Jeff's wife get = $\frac{1}{4}$ of $ x

The remaining money Jeff will have = $ x - $\frac{1}{4}$ of $ x

I.e The remaining money Jeff will have = $\frac{4 x - x}{4}$

= $\frac{3 x}{4}$

A ) The remaining amount of money is divided equally among 4 children

So, The fraction of sum of money did each child receive = $\frac{\frac{3 x}{4}}{4}$

I.e The fraction of sum of money did each child receive = $\frac{3 x}{16}$

B ) If each child will receive $ 600

∴, $\frac{3 x}{16}$ = $ 600

Or, 3 x = $ 600 × 16

Or, 3 x = $ 9600

∴ x = $\frac{9600}{3}$

I.e x = $ 3200

So, The sum of money did Jeff have $ 3200

Hence ,

A) The fraction of sum of money did each child receive is $\frac{3 x}{16}$

B) The sum of money did Jeff have $ 3200 Answer

8 0

2 years ago

What two rational expressions sum to 3x+4/x^2-6x+5?

Lubov Fominskaja [6]

Answer:

$\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}$

Step-by-step explanation:

The given rational expression is:

$\frac{3x+4}{x^{2}-6x+5} = \frac{3x+4}{(x-1)(x-5)}$

We can use concept of Partial Fractions to solve this problem. Let,

$\frac{3x+4}{(x-1)(x-5)}=\frac{A}{x-1} +\frac{B}{x-5}$

Multiplying both sides by (x - 1)(x - 5), we get:

$3x+4=A(x-5)+B(x-1)$

Substituting x = 5, we get:

$3(5)+4=A(5-5)+B(5-1)\\\\ 15+4=0+4B\\\\ 19=4B\\\\ B=\frac{19}{4}$

Substituting x = 1, we get:

$3(1)+4=A(1-5)+B(1-1)\\\\ 7=-4A\\\\ A=-\frac{7}{4}$

Substituting the value of A and B, back in the original equation, we get:

$\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}$

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2 years ago

Solve Q=3a+5ac for a

lilavasa [31]

Here is how you solve...

Q= 3a+5ac

take out the common variable.
Q= a(3+5c)

get A by its self so divide each side by (3+5c).
Q/a(3+5c)= a(3+5c)/(3+5c)

Q/a(3+5c)=A

So your answer is A= Q/a(3+5c)

hope that helped.

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