What is the measure of angle x in the figure shown?

Mathematics

1 answer:

gulaghasi [49]3 years ago

3 0

Answer:

Step-by-step explanation:

180-123=57

that line the angle is on is a straight line and a straight line is 180 degrees

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y=mx+b :)

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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The lengths of the sides of a triangle are 3, 3, and the square root 3. Can the triangle be a right triangle?

Readme [11.4K]

First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)

If the base and height were each 3, then the hypotenuse would need to be:

3^2 + 3^2 = C^2
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This is larger than sqrt(3), so this cannot be a right triangle.

If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46

This is larger than 3, this cannot be a right triangle.

The answer is b) no.

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3 years ago

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How would a find a rule for this cubic function?

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