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docker41 [41]
3 years ago
12

the scale on a blueprint for a new apartment is 1 in equals 4 feet if the length of the living room on the blueprint is 6 in, wh

at is going to be the actual length of the living room?
Mathematics
1 answer:
lianna [129]3 years ago
8 0

Answer:

<em>The actual length of the living room is 24 feet.</em>

Step-by-step explanation:

<u>Scaling</u>

The scaling on a blueprint for an apartment is:

1 inch = 4 feet

This means every 1 inch measured on the blueprint corresponds to 4 feet on the real length of the apartment.

We know the length of the living room in the blueprint is 6 inches, thus the real dimension of the living rooms is:

6 * 4 = 24 feet

The actual length of the living room is 24 feet.

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(ASAP) Billie &amp; Tony are siblings. It’s summer of 2015. Twice Billie’s age decreased by 3 is Tony’s age. Tony is fourteen ye
jeka94

Answer:

Billie's age = x = 17 years

Tony's age = y = 31 years

Step-by-step explanation:

Let

Billie's age = x

Tony's age = y

Twice Billie’s age decreased by 3 is Tony’s age.

y = 2x - 3..... Equation 1

Tony is fourteen years older than Billie.

x = y - 14

We substitute y - 14 for x in Equation 1

y = 2x - 3..... Equation 1

y = 2(y - 14) - 3

y = 2y - 28 - 3

y - 2y = -28 - 3

-y = -31

y = 31 years

Solving for x

x = y - 14

x = 31 - 14

x = 17 years.

Therefore:

Billie's age = x = 17 years

Tony's age = y = 31 years

4 0
3 years ago
Read 2 more answers
Which of the following is NOT a monomial?
Alexandra [31]

wutz da optionz????????

3 0
2 years ago
HELP MEEEEE!!! I HAVE AN F
mariarad [96]

Answer:

B is your answer!

Step-by-step explanation:

Hope I helped:)

6 0
3 years ago
Helppppp meeee pleaseeee​
ss7ja [257]
B is 0 hope thsi helps and to die lol
4 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
3 years ago
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