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2 years ago

Which are vertical angles? Check all that apply. ∠EAB and ∠BAC ∠EAB and ∠CAD ∠CAD and ∠FAE ∠CAB and ∠DAE ∠DAC and ∠DAE

Mathematics

2 answers:

Anastaziya [24]2 years ago

8 0

Answer:

<CAB and <DAE are vertical angles

<EAB and <CAD are vertical angles

Step-by-step explanation:

this is because they are formed by the prolongation of the sides of the other angles and are opposite to them on the other side of the vertex.

Rufina [12.5K]2 years ago

7 0

Answer:

A and D

Step-by-step explanation:

On ed 2020

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A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance tra

alina1380 [7]

Answer:

1.734

Step-by-step explanation:

Given that:

A local trucking company fitted a regression to relate the travel time (days) of its shipments as a function of the distance traveled (miles).

The fitted regression is Time = −7.126 + .0214 Distance

Based on a sample size n = 20

And an Estimated standard error of the slope = 0.0053

the critical value for a right-tailed test to see if the slope is positive, using ∝ = 0.05 can be computed as follows:

Let's determine the degree of freedom df = n - 1

the degree of freedom df = 20 - 2

the degree of freedom df = 18

At the level of significance ∝ = 0.05 and degree of freedom df = 18

For a right tailed test t, the critical value from the t table is :

$t_{0.05, 18} =$ 1.734

5 0

2 years ago

Using a ruler and a compass,Jimmy constructs a square inscribed within a circle. He starts by drawing a circle with a center at

lana66690 [7]

Answer:

He set the compass on a, with the width slightly wider than the distance to o and drew an arc above and below o. He then set the compass on b, without changing the width on the compass and did the same thing. True

He drew a line through where the arc pairs intersected and then labeled the points where this line crossed the top and bottom of the original circle. True

He set the compass on o, with the width slightly wider than the distance to b, and he drew a slightly larger circle around the original circle. False

He used the points that were marked along the circumference of the original circle as the vertices for the square. True

Step-by-step explanation:

Performance matters test perhaps?

5 0

2 years ago

Could 10.5\text{ cm}, 8.0\text{ cm},10.5 cm,8.0 cm,10, point, 5, start text, space, c, m, end text, comma, 8, point, 0, start te

Reptile [31]

Answer:

Yes, 10.5 cm, 8.0 cm and 4.0 cm can be the side lengths of a triangle.

Step-by-step explanation:

We have been given three lengths as 10.5 cm, 8.0 cm and 4.0 cm. We are asked to determine whether these side can be the side lengths of a triangle.

We will use triangle inequality theorem to solve our given problem.

Triangle inequality theorem states that sum of two sides of a triangle must be greater than 3rd side. Using triangle inequality theorem we will get 3 inequalities as:

$10.5+8.0>4.0$

$18.5>4.0$ True

$8.0+4.0>10.5$

$12.0>10.5$ True

$10.5+4.0>8.0$

$14.5>8.0$ True

Since all the three side lengths satisfy triangle inequality theorem, therefore, 10.5 cm, 8.0 cm and 4.0 cm can be the side lengths of a triangle.

3 0

2 years ago

Which of the following represents the zeros of f(x) = 5x3 − 6x2 − 59x + 12? (2 points)

Furkat [3]

Answer:

4, −3, 1 over 5

Step-by-step explanation:

x is a zero of $f(x)$ if $f(x) = 0$

In this problem, we have that:

$f(x) = 5x^{3} - 6x^{2} - 59x + 12$

4, 3, 1 over 5

$f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0$

So 4 is a zero of the function

$f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84$

So 3 is not a zero of the function, and this option is incorrect

4, 3, − 1 over 5

$f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84$

So 3 is not a zero of the function, and this option is incorrect

4, −3, 1 over 5

$f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0$

So 4 is a zero of the function

$f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0$

So -3 is a zero of the function

$f(\frac{1}{5}) = f(0.2) = 5*(0.2)^{3} - 6*(0.2)^{2} - 59*(0.2) + 12 = 0$

So 1 over 5 is a zero of the function

This is the correct answer.

4, −3, −1 over 5

$f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0$

So 4 is a zero of the function

$f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0$

So -3 is a zero of the function

$f(-\frac{1}{5}) = f(-0.2) = 5*(-0.2)^{3} - 6*(-0.2)^{2} - 59*(-0.2) + 12 = 23.52$

-1 over 5 is not a zero of the function

5 0

2 years ago

The image shows a geometric representation of the function f(x) = x^2 + 2x + 3 written in standard form. What is this function w

AlekseyPX

Answer:

f(x) = (x + 1)² + 2 in vertex form ⇒ 3rd answer

Step-by-step explanation:

* Lets revise how to find the vertex form the standard form

- Standard form ⇒ x² + bx + c, where a , b , c are constant

- Vertex form ⇒(x - h)² + k, where h , k are constant and (h , k) is the

vertex point (minimum or maximum) of the function

- At first we must find h and k

- By equating the two forms we can find the value of h and k

* Lets solve the problem

∵ f(x) = x² + 2x + 3 ⇒ standard form

∵ f(x) = (x - h)² + k ⇒ vertex form

- Put them equal each other

∴ x² + 2x + 3 = (x - h)² + k ⇒ open the bracket power 2

∴ x² + 2x + 3 = x² - 2hx + h² + k

- Now compare the like terms in both sides

∵ 2x = -2hx ⇒ cancel x from both sides

∴ 2 = -2h ⇒ divide both sides by -2

∴ -1 = h

∴ The value of h is -1

∵ 3 = h² + k

- Substitute the value of h

∴ 3 = (-1)² + k

∴ 3 = 1 + k ⇒ subtract 1 from both sides

∴ 2 = k

∴ The value of k = 2

- Lets substitute the value of h and k in the vertex form

∴ f(x) = (x - -1)² + 2

∴ f(x) = (x + 1)² + 2

* f(x) = (x + 1)² + 2 in vertex form

5 0

2 years ago