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alisha [4.7K]
3 years ago
9

13,000 inches equals how many miles

Mathematics
1 answer:
Nataly [62]3 years ago
8 0

Answer:

0.205177 mi

Step-by-step explanation:

Step 1: Find conversions

12 in = 1 ft

5280 ft = 1 mi

Step 2: Use Dimensional Analysis

13000 \hspace{2} in(\frac{1 \hspace{2} ft}{12 \hspace{2} in} )(\frac{1 \hspace{2} mi}{5280 \hspace{2} ft} )

0.205177 mi

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Fudgin [204]
51 ft is the answer

Explanation

The hight of the man divided by the hight of school ( which is unknown )

Equal

The length of the shadow of the man divided by the shadow of the school

Cross multiplication and find the value of unknown x which is 51... and ya
4 0
3 years ago
A cube has a volume of 64 cubic units. What is its surface area
Alona [7]
The volume of a cube is length x width x depth. In a cube all would be equal therefore. 

<span>x^3= 64 </span>
<span>x =4 </span>
<span>The area in square feet of one side would then be length x width </span>

<span>There are 6 sides so the surface area of the cube would be </span>

<span>(4)(4)(6) = 96 square feet</span>
7 0
4 years ago
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The final velocity of an object moving in one dimension is given by the formula v = x + at, where u is the
yKpoI14uk [10]

Answer:

a = (v – u)/t

Step-by-step explanation:

We know,

v = u+ at

v-u=at

(v-u) /t=a

4 0
3 years ago
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Kevin is solving this problem. 737 × 205 What are the partial products Kevin will need to solve the problem? A. 3,685 and 147,40
Andrew [12]

the answer is A

700 x 5 = 3500

30 x 5 = 150

7 x 5 = 35

3500 + 150 +35 = 3685

700 x 200 = 140,000

30 x 200 = 6000

7 x 200 = 1400

140000 + 6000 + 1400 = 147400

6 0
3 years ago
If GF is a midsegment of CDE, find CD.<br><br> A. 3.4<br><br>B. 6.8 <br><br>C. 13.6 <br><br>D. 14
bogdanovich [222]

Answer:

C


Step-by-step explanation:

Triangle CGF and triangle CED are similar. Hence, the ratio of their corresponding sides are equal. Thus we can write:

\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}

<em>We can now cross multiply and solve for CD:</em>

\frac{5x+4}{2x+3}=\frac{CD}{3x+0.8}\\(5x+4)(3x+0.8)=(2x+3)(CD)\\15x^2+4x+12x+3.2=(2x+3)(CD)\\15x^2+16x+3.2=(2x+3)(CD)\\CD=\frac{15x^2+16x+3.2}{2x+3}

Since GF is a midsegment of CDE, CD is double of CF. So we can write:

CD=2CF\\\frac{15x^2+16x+3.2}{2x+3}=2(3x+0.8)\\\frac{15x^2+16x+3.2}{2x+3}=6x+1.6\\15x^2+16x+3.2=(6x+1.6)(2x+3)\\15x^2+16x+3.2=12x^2+18x+3.2x+4.8\\15x^2+16x+3.2=12x^2+21.2x+4.8\\3x^2-5.2x-1.6=0

<em>By using quadratic formula \frac{-b+-\sqrt{b^2-4ac} }{2a} and with a=3, b= -5.2, and c= -1.6, we find the value of x to be:</em>

\frac{5.2+-\sqrt{(-5.2)^2-4(3)(-1.6)} }{2(3)}=2


<em>Since the expression for CD is \frac{15x^2+16x+3.2}{2x+3} , we plug in x=2 into this expression to find value of CD:</em>

\frac{15(2)^2+16(2)+3.2}{2(2)+3}=13.6

The correct answer is C

6 0
3 years ago
Read 2 more answers
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