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Vitek1552 [10]
2 years ago
8

A model of a car is 6.3 in. long. The scale of the model to that actual car is 1:30. What is the length of the actual car to the

nearest foot?
A. 19 ft
B. 16 ft
C. 21 ft
D. 12 ft
Mathematics
2 answers:
sergejj [24]2 years ago
8 0
8.13 is the actual legth of the car, unless i misuderstood the question.
Elan Coil [88]2 years ago
8 0

Answer:

Option B. 16\ ft

Step-by-step explanation:

we know that

The scale of the model to that actual car is equal to  \frac{1}{30}\frac{in}{in}

so

By proportion

Find the length of the actual car if the model of a car is 6.3\ in. long

\frac{1}{30}\frac{in}{in}=\frac{6.3}{x}\frac{in}{in}\\ \\x=30*6.3\\ \\x=189\ in

Convert to feet

Remember that

1\ ft=12\ in

189\ in=189/12=15.75\ ft

Round to the nearest foot

15.75\ ft=16\ ft

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Step-by-step explanation:

1. distribute 4m through the parenthesis

8mn² - 12m²n + 4m³ - 2n(5m² - 3nm) + nm(n-m)

2. use the commutative property to reorder the terms

8mn² - 12m²n + 4m³ - 2n(5m² - 3mn) + mn(n - m)

3. distribute -2n through the remaining parenthesis

8mn² - 12m²n + 4m³ -10m²n + 6mn² + mn² - m²n

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8mn² + 6mn² + mn² - 12m²n - 10m²n - m²n + 4m³

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15mn² - 23m²n +4m³

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15mn² -23m²n + 4m³

5 0
3 years ago
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Step-by-step explanation:

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4 0
3 years ago
Find the average value of f over the region D. f(x, y) = 3xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 9)
MakcuM [25]

The average value of f over the region D is 243/4

To answer the question, we need to know what the average value of a function is

<h3>What is the average value of a function?</h3>

The average value of a function f(x) over an interval [a,b] is given by

\frac{1}{b - a} \int\limits^b_a {f(x)} \, dx

Now, given that we require the average value of f(x,y) = 3xy over the region D where D is the triangle with vertices (0, 0), (1, 0), and (1, 9).

x is intergrated from x = 0 to 1 and the interval is [0,1] and y is integrated from y = 0 to y = 9

So, \frac{1}{b - a} \int\limits^b_a {f(x,y)} \, dA = \frac{1}{1 - 0} \int\limits^1_0 \int\limits^9_0 {3xy} \, dxdy \\= \frac{3}{1} \int\limits^1_0 {x} \,dx\int\limits^9_0 {y} \,dy\\ =  \frac{3}{1} [\frac{x^{2} }{2} ]^{1}_{0}[\frac{y^{2} }{2} ]^{9}_{0}  \\= 3[\frac{1^{2} }{2} - \frac{0^{2}}{2} ] [\frac{9^{2} }{2} - \frac{0^{2}}{2} ] \\= 3[\frac{1}{2} - 0 ][\frac{81}{2} - 0 ]\\=  \frac{81}{2} X3 X \frac{1}{2} \\=  \frac{243}{4}

So, the average value of f over the region D is 243/4

Learn more about average value of a function here:

brainly.com/question/15870615

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8 0
1 year ago
Show the processes for solving 2x+3y=5 and 4x - y=17 using elimination and substitution
r-ruslan [8.4K]

<u>ANSWER:  </u>

The solution of the two equations 2x+3y=5 and 4x - y=17 is (4, -1).

<u>SOLUTION: </u>

Given, two linear equations are 2x + 3y = 5 → (1) and 4x – y = 17 → (2).

Let us first solve the above equations using <em>elimination process. </em>

For elimination, one of the coefficients of variables has to be same in order to cancel them.

Now solve (1) and (2)

eqn (1) \times 2 → 4x + 6y = 10

eqn (2) → 4x – y = 17

(-) ----------------------------

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y = -1

Substitute y value in (2)

4 x-(-1)=17 \rightarrow 4 x+1=17 \rightarrow 4 x=17-1 \rightarrow 4 x=16 \rightarrow x=4

So, solution of two equations is (4, -1).

<u><em>Now let us solve using substitution process.</em></u>

Then, (2) → 4x – y = 17 → 4x = 17 + y → y = 4x – 17

Now substitute y value in (1) → 2x + 3(4x – 17) = 5 → 2x + 12x – 51 = 5 → 14x = 5 + 51 → 14x = 56  

x = 4

Substitute x value in (2) → y = 4(4) – 17 → y = 16 – 17 → y = -1

Hence, the solution of the two equations is (4, -1).

7 0
2 years ago
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