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3 years ago

Can I get Help on these questions ???

Mathematics

1 answer:

iragen [17]3 years ago

8 0

∠A must be 2*22=44 because it is made up of two times y.
A triangle's corners should add up to 180, so B and C together are 180-44=136. Divide 136 equally between B and C due to symmetry, leaves 136/2 = 68° for C

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3/4=x/6 proportion please answer

mina [271]

Answer:

The answer is 9/2

Step-by-step explanation:

3/4 = x/6

or, 4x = 18

or, x = 18/4

or, x = 9/2

x = 9/2

5 0

3 years ago

What is the slope of the line given by the equation below? y = 5x - 6

Bogdan [553]

Slope-intercept form is y = mx + b, where m is the slope of the line and b is the y-intercept.
For the equation y = 5x - 6, m = 5, meaning that the slope of the line is 5.

4 0

3 years ago

Read 2 more answers

How to do this ....?

ICE Princess25 [194]

I don’t know what you’re learning so tell me and I’ll give you what I know about it.

7 0

3 years ago

1609 A C B. What is the mZABC ?

Free_Kalibri [48]

Answer:

< ABC = 80

Step-by-step explanation:

Inscribed Angle = 1/2 Intercepted Arc

< ABC = 1/2 ( 160)

< ABC = 80

3 0

2 years ago

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0

3 years ago