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ryzh [129]
3 years ago
13

Kim is drawing a map of the different schools in her school district. She knows that her middle school is 10.6 miles away from t

he middle school that her best friend attends. If every 2 inches on the map represents 5 miles, how far apart will the two schools be on the map, to the nearest tenth of an inch
Mathematics
2 answers:
Stolb23 [73]3 years ago
8 0
4 inches is the answer
mamaluj [8]3 years ago
3 0
4 inches your welcome
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Pedro Pascal wants to have $15,000 in an account after five years. He found a bank that will offer him a 7.5% interest rate, com
Juli2301 [7.4K]

<u>Answer-</u>

He needs to put $10345 in the account today, in order to get 15000 in five years.

<u>Solution-</u>

We know that,

A=P(1+\frac{r}{n})^{nt}

Where,

A = future value of the investment with interest  = 15000

P = principal investment amount

r = annual interest rate (decimal)  = 7.5% = 0.075

n = number of times that interest is compounded per year  = 4

t = the number of years the money is invested = 5

Putting the values,

15000=P(1+\frac{0.075}{4})^{4 \times 5}

\Rightarrow P=\frac{15000}{(1+0.01875)^{20}} =\frac{15000}{1.01875^{20}} =\frac{15000}{1.45} =10344.8 \approx 10345

6 0
3 years ago
A cereal box is an example of a
JulsSmile [24]

Answer: recantagle

Step-by-step explanation:

3 0
3 years ago
Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16
4vir4ik [10]

Answer:

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

9 -16 (x^2 + y^2) = 0 \\ \\  16 (x^2 + y^2)  = 9 \\ \\  x^2+y^2 = \dfrac{9}{16}

x^2+y^2 = (\dfrac{3}{4})^2

where:

0 ≤ r ≤ \dfrac{3}{4} and 0 ≤ θ ≤ 2π

∴

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0}  \ \ \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0}  ( 9r^2-16r^4})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

4 0
3 years ago
H varies directly as L. If H=20 when L=50, determine H when L=30
vichka [17]

The correct answer is 12

Set up a ratio and then solve. See paper attached. (:

4 0
4 years ago
A box contains 8 $1’s 14 $5’s 9 $10’s 15 $20’s 2 $50’s and 7 $100’s
Serjik [45]
The total money is $2,178
5 0
3 years ago
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