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4 years ago

Find the quotient 130 ÷ –13. A. –12 B. –9 C. –10 D. –5

Mathematics

2 answers:

WITCHER [35]4 years ago

3 0

Answer:

C)-10

Step-by-step explanation:

negative divided by negative is positive.

-130 / -13 =10

Drupady [299]4 years ago

3 0

Answer:

Option C, -10

Step-by-step explanation:

130 / -13

13 goes into 130 ten times

-10

Answer: Option C, -10

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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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