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lidiya [134]
3 years ago
8

Perpendicular and Angle bisectors​

Mathematics
1 answer:
Stels [109]3 years ago
3 0
The answer is 8, because CA and BC are equal
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0.2x + (-0.9) + 1.7 = 9.6<br> 0.2x + 0.8 = 9.6
user100 [1]

Answer:44

Step-by-step explanation:

0.2x+(-0.9)+1.7==9.6

0.2x+0.8=9.6

Subtract 0.8 from both sides

0.2x+0.8-0.8=9.6-0.8

0.2x=8.8

Divide both sides by 0.2

0.2x ➗ 0.2=8.8 ➗ 0.2

x=44

6 0
3 years ago
Find the slope of the line containing the points (6,4) and (6,8)
kramer

Answer:

Undefined.

Step-by-step explanation:

Note that the x-coordinate does not change.  This indicates that the two points lie on a vertical line.  The slope of a vertical line is undefined.


6 0
3 years ago
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What fraction of a dollar would this represent in lowest terms
Elza [17]
I think 1/4 is the answer
3 0
4 years ago
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he graph below represents the price of sending video messages using the services of a phone company. What is the constant of pro
Colt1911 [192]

Answer:

24.

Step-by-step explanation:

Basically constant of proportionality is a ratio.

We take two points, we'll take the order pair (2,48)

It will always be y/x so y is 48 and x is 2

48/2    -  now we simplify

answer: 24

hope this helps! :))

8 0
3 years ago
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From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends
valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let,  AB = y, DC = x.

In Δ ABC,

\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}

Similarly, in Δ ACD,

\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft

Hence, the mat is 4.33 ft high off the ground.

5 0
3 years ago
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