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4 years ago

Simplify the expression. 4(6 – 3) a. 7 b. 12 c. 21 d. 24

Mathematics

1 answer:

goldfiish [28.3K]4 years ago

6 0

B. 12 is the awnser.

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Original 48.60 markdown 100%

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Insufficient info!!

But if the original cost is $48.60, and you mark it down 100%, then you are essentially giving the item away free.

$48.60 less 100% of $48.60 is $0.

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3 years ago

Which is an example of the opposite of a sum property

Komok [63]

C is the answer to this problem

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kondor19780726 [428]

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3 years ago

In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd2, find ABMC, AAMD, and ACMD.

Komok [63]

Since point M is the midpoint of AB, then AM=MB.

Consider the area of the triangles ABC and BMC:

$A_{ABC}=\dfrac{1}{2}\cdot AB\cdot h_c=56\ yd^2,$

where $h_c$ is the height drawn from the vertex C to the side AB.

So, $AB\cdot h_c=112\ yd^2.$

Now

$A_{BMC}=\dfrac{1}{2}\cdot BM\cdot h_c=\dfrac{1}{2}\cdot \dfrac{AB}{2}\cdot h_c=\dfrac{1}{4}\cdot AB\cdot h_c=\dfrac{1}{4}\cdot 112=28\ yd^2.$

Also

$A_{AMC}=A_{ABC}-A_{BMC}=56-28=28\ yd^2.$

Now consider the area of the triangles AMD and CMD. Let $h_M$ be the height drawn from the point M to the side AC.

$A_{AMD}=\dfrac{1}{2}\cdot AD\cdot h_M=\dfrac{1}{2}\cdot \dfrac{2AC}{7}\cdot h_M=\dfrac{2}{7}\cdot \left(\dfrac{1}{2}\cdot AC\cdot h_M\right)=\dfrac{2}{7}\cdot A_{AMC}=\dfrac{2}{7}\cdot 28=8\ yd^2.$

Therefore,

$A_{MDC}=A_{AMC}-A_{AMD}=28-8=20\ yd^2.$

Answer: $A_{MBC}=28\ yd^2, A_{AMD}=8\ yd^2, A_{MDC}=20\ yd^2.$

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3 years ago

Read 2 more answers

Find the lost common denominator of 3/4 and 5/8

Anika [276]

I assume you mean least. That would be 8.

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3 years ago