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DENIUS [597]
4 years ago
5

A child is given an allowance of $125 per day for chores. The parents have promised to increase it by $0.75 per day after a mont

h. What is the percent increase? ********
Mathematics
1 answer:
meriva4 years ago
5 0
1. 125x = 125.75

2. 125x/125 = 125.75/125

3. x= 0.01006%

ANSWER: It increases 0.01006% each day.
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I am Lyosha [343]
Yes if x=5 then 4(5)= 20 and 15+5=20
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3 years ago
Explain your answer !! <br> Have a nice day <br><br> Will give braisnlt
11111nata11111 [884]

Answer:

The price of hot dog is $2.5 and that of a pretzel is $1.25.

Step-by-step explanation:

Let the cost of hotdogs be x and that of pretzels be x.

So,

4x+3y = 13.75 ...(1)

and

2x+5y = 11.25 ...(2)

Multiply equation (2) by 2.

4x+10y = 22.5 ...(3)

Subtract equation (1) from (3).

4x+10y-(4x+3y) = 22.5-13.75

4x+10y-4x-3y = 8.75

7y = 8.75

y = 1.25

Put the value of y in equation (1).

4x+3(1.25) = 13.75

4x+3.75 = 13.75

4x = 13.75-3.75

4x = 10

x = 2.5

So, the price of hot dog is $2.5 and that of a pretzel is $1.25.

3 0
3 years ago
Share $600 between Anna and Raman in the ratio 3:7 and 1:4
Ulleksa [173]

Answer:

3 : 7 =>

Anna = $180

Raman = $420

1 : 4 =>

Anna = $120

Raman = $480

Step-by-step explanation:

<u>Shares when ratio is 3 : 7</u>

Anna : Raman = 3 : 7

Sum of ratio 10

Anna's share=

                   \frac{3}{10} \times 600 = \$  180

Raman 's share=

                  \frac{7}{10} \times 600 = \$ 420

<u>Shares when ratio is 1 : 4</u>

Anna : Raman = 1 : 4

Sum of ratio = 5

Anna 's share =

                   \frac{1}{5} \times 600 = \$ 120

Raman's share =

                  \frac{4}{5} \times 600 = \$ 480

8 0
3 years ago
GJ bisects ∠FGH and is a perpendicular bisector of FH. What is true of triangle FGH?
professor190 [17]
Referring to the image attached:

this is an isosceles triangle

6 0
3 years ago
Read 2 more answers
Find the coefficient of the x^3 in the expansion of (2x-9)^5
il63 [147K]

Use the binomial theorem:

\displaystyle (2x-9)^5 = \sum_{k=0}^5 \binom5k (2x)^{5-k}(-9)^k = \sum_{k=0}^5 \frac{5!}{k!(5-k)!} 2^5 \left(-\frac92\right)^k x^{5-k}

The <em>x</em> ³ terms occurs for 5 - <em>k</em> = 3, or <em>k</em> = 2, and its coefficient would be

\dfrac{5!}{2!(5-2)!} 2^5 \left(-\dfrac92\right)^2 = \boxed{6480}

8 0
3 years ago
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