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3 years ago

If each 2 cm on the scale equals 8 feet what are the actual dimensions of the room? (1 point) width=4 length= 6

Mathematics

1 answer:

Shalnov [3]3 years ago

3 0

The actual dimensions are 16 feet wide and 24 ft long. And 16 x 24 is 384.

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If the side length is 6. cm, then the area is ______cm2 ?

ArbitrLikvidat [17]

Answer:

36 cm2

Step-by-step explanation:

tep-by-Step:

Start with the formula:

Area = a2

Don't forget: a2 = a × a (a squared).

Substitute the side length into the formula. In our example, a = 6.

Area = 6^2 = 6 × 6 = 36 cm2

Answer:

The area of the square with sides of length 6 cm is 36 cm2.

8 0

3 years ago

WILL MARK BRAINLIEST An architect is drafting a scale drawing of a house for an upcoming project. The customer wants the front o

kobusy [5.1K]

Answer:

60 ft is 720 in. I did 720/15...that equals 48. At this point i think each in in the drawing is a foot, so every one in is 48 ft.

(x+48)

there is no y because they only give information about one side. hope this helps. :)

Step-by-step explanation:

4 0

3 years ago

1. Integers are not closed under which operation?

velikii [3]

Answer:

Integers are not closed under the DIVISION operation

Example:

(9 ÷ 2 = 4½)

4 0

3 years ago

T F IfA and B are similar matrices, then AT=BT

muminat

Answer:

Step-by-step explanation:

We know that for two similar matrices $A$ and $B$ exists an invertible matrix $P$ for which

$[tex]B = P^{-1} AP$ [/tex]

∴ $B^{T} = (P^{-1})^{T} A^{T} P^{T} \\$

Also $P^{-1}P = I\\$

and $(P^{-1})^{T} = (P^{T})^{-1}$

∴ $(P^{-1})^{T}P^{T} = I$

so, $B^{T} = (P^{-1})^{T} A^{T} P^{T} = (P^{T})^{-1}A^{T} P^{T}\\B^{T} = A^{T} I\\B^{T} = A^{T}$

7 0

3 years ago

A gun mass of 5 kg fired a bullet of mass 10 g with the velocity of 360 km/h. What

klemol [59]

Answer:

The recoil velocity of the gun is $0.72\,\,\frac{km}{h}$ and is pointing in opposite direction to the velocity of the bullet.

Step-by-step explanation:

Use conservation of linear momentum, which states that the momentum of the bullet (product of the bullet's mass times its speed) should equal in absolute value the momentum of the recoiling gun (its mass times its recoil velocity).

We also write the mass of the bullet in the same units as the mass of the gun (for example kilograms). Mass of the bullet = 0.010 kg

In mathematical terms, we have:

$5\, kg * v= 0.01 \,kg\,* 360\,\frac{km}{h} \\v=\frac{0.01\,*360}{5} \,\,\frac{km}{h}\\v=0.72\,\,\frac{km}{h}$

6 0

3 years ago