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3 years ago

10

Sara packed 7 boxes of books. Each box had 13 mystery books and 15 science fiction books. How many more science fiction books th

an mystery books did Sara pack?

1 answer:

Mrrafil [7]3 years ago

5 0

Answer:

There are 14 more science fiction books than mystery books

Step-by-step explanation:

Multiply 13 x 7 = 91

Multiply 15 x 7 = 105

Subtract 105 - 91 = 14

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What is the value of y?<br><br><br><br> Enter your answer in the box.

Elenna [48]

The value of y = 40.

5x = 40
x = 8

180 = 40 + 40 + (2y + 20)
40 = y

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3 years ago

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Tim's company offers a reimbursement package of $0.65 per mile plus $145

zalisa [80]

Answer:

C = 0.65x + 145

Step-by-step explanation:

Assume:

Number of miles = x

Annual maintenance = $145

Total reimbursement = C = $0.65*x + $145 = $0.65x + $145

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3 years ago

Consider a home mortgage of $225,000 at a fixed APR of 4.5 % for 30 years. a. Calculate the monthly payment. b. Determine the

White raven [17]

<u>The formula for the monthly payment</u>.....

$M= \frac{P(1+r)^nr}{[(1+r)^n] -1}$ , where P = the principal amount, r = the monthly interest rate and n = the total number of months.

Here annual interest rate is given as 4.5%

So, the monthly interest rate $=\frac{4.5\%}{12}= 0.375\% = 0.00375$

Total number of months $= (30*12)months=360 months$

Also given that, the principal amount is $225000

a. So, the monthly payment will be.....

$M= \frac{P(1+r)^nr}{[(1+r)^n] -1}\\ \\ M= \frac{225000(1+0.00375)^3^6^0*0.00375}{(1+0.00375)^3^6^0 -1}\\ \\ M= \frac{225000(1.00375)^3^6^0*0.00375}{(1.00375)^3^6^0 -1} \\ \\ M \approx 1140$

Thus, the monthly payment will be approximately $1140

b. The <u>total amount paid</u> over the term of the loan will be: $\$ 1140*360 months = \$ 410400$

c. As the principal amount was $225000 , so the amount of interest $= (\$ 410400- \$ 225000)= \$ 185400$

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3 years ago

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Mashutka [201]

Answer:

$\sqrt\frac{387}{20}$

Step-by-step explanation:

$Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx$

$y=\dfrac{3}{5}x^{\frac{5}{3}}- \dfrac{3}{4}x^{\frac{1}{3}}+6$

$\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}$

$1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})$

$=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}$

For the Interval $1\leq x\leq 8$

Length of the Curve $=\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\$

Using T1-Calculator

$=\sqrt\frac{387}{20}$

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3 years ago

Someone plz help meee

coldgirl [10]

Answer:

The second one

Step-by-step explanation:

7 0

3 years ago