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3 years ago

I don’t know how to solve for m/ the slope and b/ the y-intercept on a horizontal line.

Mathematics

1 answer:

pav-90 [236]3 years ago

8 0

Answer:

the y-intercept is where u start on a graph like where it crosses through the middle line and the slope it the rise and run so rise is how much it goes up and run is how much u go over.

hope this helps.

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In ΔEFG, the measure of ∠G=90°, GE = 32 feet, and EF = 41 feet. Find the measure of ∠E to the nearest tenth of a degree.

USPshnik [31]

Answer:

m∠E = 38.7°

Step-by-step explanation:

If you make a drawing it will help. What you have is a right triangle where you are given the adjacent side to ∠E and the hypotenuse.

The cosine of a angle = adjacent side/hypotenuse.

Therefore cos E = 32/41

arccos (32/41) = 38.7°

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4 years ago

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sammy [17]

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3 years ago

Find the value that makes the ratios equal ? to 12, 5 to 4 <br> A. 25<br> B. 15<br> C. 20 <br> D. 16

Nonamiya [84]

Answer:

B.15

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3 years ago

In the diagram, AD=CD=CB and measure of angle A=40°. How many degrees are in angle DCB?

Ad libitum [116K]

Answer:

$20^{\circ}$

Step-by-step explanation:

Consider triangle ACD. This triangle is isosceles triangle, because AD = DC. Angles adjacent to the base AC are congruent abgles, so

$m\angle DAC=m\angle DCA=40^{\circ}$

The sum of the measures of all interiror angles in the triangle is always $180^{\circ},$ then

$m\angle DAC+m\angle DCA+m\angle ADC=180^{\circ}\\ \\40^{\circ}+40^{\circ}+m\angle ADC=180^{\circ}\\ \\m\angle ADC=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}$

Angles ADC and BDC are supplementary angles (add up to $180^{\circ}$ ), then

$m\angle BDC=180^{\circ}-100^{\circ}=80^{\circ}$

Consider triangle BCD. This triangle is isosceles triangle because BC = DC. Angles adjacent to the base BD are congruent abgles, so

$m\angle BDC=m\angle DBC=80^{\circ}$

The sum of the measures of all interiror angles in the triangle is always $180^{\circ},$ then

$m\angle DBC+m\angle BDC+m\angle BCD=180^{\circ}\\ \\80^{\circ}+80^{\circ}+m\angle BCD=180^{\circ}\\ \\m\angle BCD=180^{\circ}-80^{\circ}-80^{\circ}=20^{\circ}$

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3 years ago