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3 years ago

Could anyone help me with these two questions? I need the answer by November 4th

Mathematics

1 answer:

TiliK225 [7]3 years ago

7 0

Answer:

Answer for question 2: A

Answer for question 3: -0.728

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What is the y= form of x+y=1?

Ulleksa [173]

Answer:

y=1-x

Step-by-step explanation:

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3 years ago

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8000 + 8800 + 8000 + 8 million + 80 trillion + another a katrillion equals

Yuri [45]

Answer:

8,000,802,4800

Step-by-step explanation:

Katrillion isn't a real value but everything else is.

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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

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Can you guys help please!! Due very soon!

natali 33 [55]

the y-intercept is at (0,2) where it's touching the y-axis. That's why it's called y-intercept.

Hope thishelps!

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3 years ago

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PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPP

sineoko [7]

Answer:

I'm pretty sure that it is A- The message travels through the spinal cord

Step-by-step explanation:

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3 years ago