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2 years ago

14

Which equation, in slope-intercept form, matches the equation shown?

1 answer:

max2010maxim [7]2 years ago

6 0

Answer: i think it is y=−56x−4

Step-by-step explanation:

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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

3 years ago

154 - 4(1 + 6m) + 6m

polet [3.4K]

Answer:

6(25 - 3M)

Thats the correct answer

5 0

3 years ago

Im timed make it quick

erica [24]

Answer:

i think its a

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Lee filled up his gas tank before starting a road trip. If he has used 5.8 gallons two hours into the trip, and has 12.4 gallons

KATRIN_1 [288]

He started out with 19.1 gallons because if you add 5.8 and 12.4 you get 19.1

3 0

3 years ago

Prove the divisibility:<br><br>45^45·15^15 by 75^30

garri49 [273]

Answer:

$3^{75}$ .

Step-by-step explanation:

We have been an division problem: $\frac{45^{45}*15^{15}}{75^{30}}$ .

We will simplify our division problem using rules of exponents.

Using product rule of exponents $(a*b)^n=a^n*b^n$ we can write:

$45^{45}=(9*5)^{45}=9^{45}*5^{45}$

$15^{15}=(3*5)^{15}=3^{15}*5^{15}$

$75^{30}=(15*5)^{30}=15^{30}*5^{30}$

Substituting these values in our division problem we will get,

$\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}$

$\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}$

Using product rule of exponents $(a*b)^n=a^n*b^n$ we will get,

$\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}$

$\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}$

Using power rule of exponents $a^n*a^m=a^{n+m}$ we will get,

$\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}$

$\frac{3^{105}*5^{60}}{3^{30}*5^{60}}$

$\frac{3^{105}}{3^{30}}$

Using quotient rule of exponent $\frac{a^m}{a^n}=a^{m-n}$ we will get,

$\frac{3^{105}}{3^{30}}=3^{105-30}$

$3^{105-30}=3^{75}$

Therefore, our resulting quotient will be $3^{75}$ .

7 0

3 years ago