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3 years ago

True or false please help

Mathematics

1 answer:

GarryVolchara [31]3 years ago

4 0

True because if you did 180°-270° it would be in quadrant 3 or it could be on quadrant 2 because quadrant 2 is 90-180

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How do you turn -3.625 into a fraction.

noname [10]

-3.625

=-(3000/1000)-(625/1000)

=-3625/1000

=-725/200

=-145/40

=-29/8

5 0

4 years ago

Read 2 more answers

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

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Can u pls help me select the right answer^.^

Gennadij [26K]

Step-by-step explanation:

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3 years ago

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Please help me Solve P=2x+2w for x thank you

Lilit [14]

P = 2(x + w)
P/2 = w + x
x = P/2 - w

4 0

3 years ago

∠3 and ∠4 are supplementary angles. If ∠3 measures 50°, what is the measure of ∠4

Sidana [21]

Answer:

130

Step-by-step explanation:

Supplementary angles are angles that are equal to 180 degree so if one is 50 degrees then the other must be added to it and get 180

180= 50 + ∠4

-50 -50

130=∠4

6 0

3 years ago