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3 years ago

Simplify 1/18 / 1/6

Mathematics

1 answer:

mariarad [96]3 years ago

8 0

The equation in the question is not the same as the one in the picture so here's the answers to both of them.

Q1 - 1/3 or 0.333333

Q2 - 1/108 or 0.009259

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Type the expression that results from the following series of steps:

alexira [117]

Answer:

7y - 2

Since two is not able to subtract, just leave it.

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3 years ago

If an angle is 10 degree less than its complement, find the angle.<br>

GrogVix [38]

Let one be x

Other one is x-10

ATQ

$\\ \sf \longmapsto x+x-10=90$

$\\ \sf \longmapsto 2x-10=90$

$\\ \sf \longmapsto 2x=90+10$

$\\ \sf \longmapsto 2x=100$

$\\ \sf \longmapsto x=\dfrac{100}{2}$

$\\ \sf \longmapsto x=50$

$\\ \sf \longmapsto x-10=50-10=40$

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3 years ago

It was estimated that the United States would need 6200 more veterans in 2012. The percent error was 10%. If the estimate was lo

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4 years ago

Don't forget! If you multiply or divide by a negative number, you must

LUCKY_DIMON [66]

Answer:

c > 20

make a number line that starts with the number 16 and ends with 22.

on the number 20 circle it (don't fill in the circle).

then draw an arrow to the right on top of the circle that you circled.

Step-by-step explanation:

7 0

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

5 0

3 years ago

Read 2 more answers

Simplify 1/18 / 1/6​

Simplify 1/18 / 1/6