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3 years ago

Write an expression in simplest form for the area of rectangle with length 8c meters and width 2 meters

Mathematics

1 answer:

DIA [1.3K]3 years ago

6 0

8•2. The area is length times width

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Taking away 5tens from 400gives

RideAnS [48]

Answer:

350

Step-by-step explanation:

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2 years ago

If the ramp will be 3 feet above the ground and have a 4 foot Base , How long will the ramp be?

FinnZ [79.3K]

Answer:

5 feet long

Step-by-step explanation:

According to the Pythagorean Theorem, a^2 + b^2 = c^2 where "a" and "b" are the two legs of the right triangle and "c" is the hypotenuse.

We are trying to find the part that is at an angle, which is "c". So if we know "a" is "3 and "b" is 4, we can plug the values into the equation.

3^2 + 4^2 = c^2 plug in the values

9 + 16 = c^2 simply the two squared values

25 = c^2 add the two values

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6 0

3 years ago

Read 2 more answers

EASY BRAINLIEST PLEASE HELP!!

butalik [34]

Answer:

see below

Step-by-step explanation:

<h3>Proposition:</h3>

Let the diagonals AC and BD of the Parallelogram ABCD intercept at E. It is required to prove AE=CE and DE=BE

<h3>Proof:</h3>

1)The lines AD and BC are parallel and AC their transversal therefore,

$\displaystyle \angle DAC = \angle ACB \\ \ \qquad [\text{ alternate angles theorem}]$

2)The lines AB and DC are parallel and BD their transversal therefore,

$\displaystyle \angle BD C= \angle ABD \\ \ \qquad [\text{ alternate angles theorem}]$

3)now in triangle ∆AEB and ∆CED

$\displaystyle \angle EAD=\angle ECB$
$\angle EDA=\angle EBC$
$\displaystyle AD=BC$

therefore,

$\displaystyle \Delta AEB \cong \Delta CED$

hence,

AE=CE
DE=BE

Proven

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3 years ago

0.0000713 converted to scientific notation is

viva [34]

Answer:

7.13 x 10^-5

Step-by-step explanation:

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3 years ago

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Maria drew two parallel lines kl and mn intersected by a transversal pq, as shown below: two parallel lines kl and mn with pq as

hjlf

${\text{Here, two parallel lines }}kl\,{\text{and }}mn\,{\text{are intersected by a}} \hfill \\ {\text{transversal }}pq\,{\text{at points }}r\,{\text{and }}s\,{\text{respectively}}{\text{. }} \hfill \\ {\text{Here, two angles }}krq\,{\text{and}}\,msq\,{\text{formed}}\,{\text{by the transversal }}pq\,{\text{with respect to the lines}} \hfill \\kl\,{\text{and }}mn\,{\text{such that}} \hfill \\ m\left( {\angle \;krq} \right)\, = \,m\left( {\angle \;msq} \right) \hfill \\text{as these two angles are corresponding angles which are always of equal measure,}} \hfill \\ {\text{i}}{\text{.e}}{\text{., corresponding angles are always congruent to each other}}{\text{.}} \hfill \\$

4 0

3 years ago

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