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3 years ago

For the diagram shown, which angles are alternate interior angles? ∠3 and ∠6 ∠8 and ∠2 ∠4 and ∠7 ∠3 and ∠5

Mathematics

2 answers:

labwork [276]3 years ago

8 0

Answer:

∠3 and ∠5 are alternate interior angles

Step-by-step explanation:

By the definition, alternate interior angles are angles that are inside the lines and opposite sides of each other on the transversal.

Karolina [17]3 years ago

6 0

Answer:

3 and 5

Step-by-step explanation:

Alternate interior angles are the angles in between the two lines and on the inside

4 and 6

3 and 5

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Find the slope of the line passing through the points (-3, -8) and (5, 6).

Ulleksa [173]

Answer:

7/4 or 7 over 4

Step-by-step explanation:

slope formula: (y2 - y1) / (x2 - x1)

= (6 - (-8)) / (5 - (-3))

= (6 + 8) / (5 + 3)

= 14 / 8

=7 / 4

6 0

3 years ago

Read 2 more answers

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago

PLEASE ANSWER WILL MARK BRAINLIEST!!! What is the slope of y = -4x – 3?

natali 33 [55]

I think -4
Because the form is y=ax + b
a is the slop so -4 is the answer.

4 0

3 years ago

What is the area of the square of its sides measure 2 2/3m

larisa [96]

Answer:

A = 7 1/9 m^2

Step-by-step explanation:

The area of a square is the side length squared.

A = s^2

A = (2 2/3) ^2

Change the mixed number to an improper fraction

2 2/3 = (3*2+2) /3 = 8/3

A = (8/3) ^2

= 64/9

Change this back to a mixed number

9 goes into 64 seven times (9*7 = 63 with 1 left over)

7 1/9 m^2

The area is 7 1/9

5 0

2 years ago

What conditions would be enough to prove that P is the circumcenter of HJK ?

BlackZzzverrR [31]

Answer:

Option A and C

Step-by-step explanation:

Explanation for why I choose option A :-

If you look at the triangle, HK has a midpoint of point L, HJ has a midpoint of point M, and KJ has a midpoint of point N. Now if you stretch the lines inside of the triangle, they intersect at point P which is known as the circumcenter of HJK, true.

Explanation for why I choose option C :-

Now let us look at the lines stretch inside of the triangle. There are three lines which are congruent. Those lines are PK, HP, and PJ (they are the big lines inside the triangle) are congruent. Since those three line intersect at point P, it make the statement true that P is the circumcenter of HJK.

Why option B is not an answer?

We do not know the third line (PN) measurements to find that if the three lines (PL, PM, and PN) are congruent and equal 90⁰. Since we do not know, it would not prove that they are congruent and P is the circumcenter of HJK.

Why option D is not an answer?

Well it says that triangle HJK is an acute triangle which is true, but this doesn't prove that P is the circumcenter of HJK.

I hope this helps, thank you :) !!

4 0

3 years ago