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The figure below shows a straight line AB intersected by another straight line t:

BaLLatris [955]

When two straight lines intersect, the pairs of nonadjacent angles in opposite posi-tions are known as vertical angles.

If a segment AB is intersected by a transversal labeled t, then ∠1 and ∠3 and ∠2 and ∠4 are vertically angles formed by the transversal t on the segment AB.

Angles ∠1 and ∠2 can be described as adjacent and supplementary angles, so

$m\angle 1+m\angle 2=180^{\circ}$ .

Angles ∠3 and ∠2 can be also described as adjacent and supplementary angles, so

$m\angle 3+m\angle 2=180^{\circ}$ .

Subtract from the first equation the second equation:

$m\angle 1+m\angle 2-(m\angle 3+m\angle 2)=180^{\circ}-180^{\circ},\\ m\angle 1-m\angle 3=0,\\ m\angle 1=m\angle 3$ .

Similarly you can prove that $m\angle 2=m\angle 4$ .

4 0

3 years ago

I don't get this one plz help

ololo11 [35]

Isocoleese means 2 angles are same, normally the base angles

so since it says they measure 42
rememeber that all angles in traingle add to 180
so base angles+vertex angle=180
base angles aer both 42 so
2 times 42+vertex angle=180
42 times 2=84
so
if veretx angle=z then
84+z=180

answer is B

3 0

3 years ago

Read 2 more answers

Explain how you would estimate to subtract 189 from 643

Dima020 [189]

I would just round up to 200 and then subtract twelve or just subtract 200 what ever way u prefer.

5 0

3 years ago

If f(x)=2x^3-6x^2-16x-20f(x)=2x *3 −6x *2−16x−20 and f(5)=0, then find all of the zeros of f(x)f(x) algebraically.

mihalych1998 [28]

The zeros of the cubic function f(x) = 2x³ - 6x² - 16x - 20 are given as follows:

x = 5, x = -1 + i, x = -1 - i.

<h3>How to obtain the solutions to the equation?</h3>

The equation is defined by the rule presented as follows:

f(x) = 2x³ - 6x² - 16x - 20.

One solution for the equation is given as follows:

x = 5.

Because f(5) = 0.

Then (x - 5) is a linear factor of the function f(x), which can be written as follows:

2x³ - 6x² - 16x - 20 = (ax² + bx + c)(x - 5).

This is because the product of a linear function and a quadratic function results in a cubic function.

Now we expand the right side to begin finding the coefficients of the quadratic function that we are going to solve to find the remaining zeros:

2x³ - 6x² - 16x - 20 = = ax³ + (b - 5a)x² + (c - 5b)x - 5c.

Then these coefficients are obtained comparing the left and the right side of the equality as follows:

a = 2.

-5c = -20 -> c = 4.

b = -6 + 5a = 4.

Hence the equation is:

2x² + 4x + 4.

Using a quadratic equation calculator, the remaining zeros are given as follows:

x = -1 + i.

x = -1 - i.

More can be learned about the solutions of an equation at brainly.com/question/25896797

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8 0

1 year ago

the strength of students in a school when increased by 8% becomes 2160. find the original strength of students in the school.

Leni [432]

Answer:

1987.2

Step-by-step explanation:

We can write a proportion to find the total amount in the choir using the information given. A proportion is two equivalent ratios set equal to each other. Since we know the strength of students increased by 8% to 2160 then the original was 92%.

$\frac{92}{100}=\frac{y}{2160}$

We will cross multiply the numerator of one ratio with denominator of the other. And then solve for y.

92(2160)=100(y)

198720=100y

y=1987.2

There are 1987.2 as the original strength of students.

6 0

3 years ago

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