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2 years ago

The function y = f(x) is graphed below. What is the average rate of change of the

Mathematics

1 answer:

Yuliya22 [10]2 years ago

7 0

Don’t know but hope you have a good day

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54 ft = ____ yd 9. 11 L ≈ ____ qt 10. 80 g = ____ kg <br><br>i need help with this

Alexxandr [17]

Answer:

54 ft = 18 yd 9. 11 L ≈ 11.6236 qt 10. 80 g = 0.08 kg

Step-by-step explanation:

3 0

3 years ago

What could be the next statement that Ivan makes his proof?

natali 33 [55]

Answer: Choice A) Triangle ABC is similar to triangle ACD by AA

AA stands for Angle Angle. Specifically it means we need 2 pairs of congruent angles between the two triangles in order to prove the triangles similar. Your book might write "AA similarity" instead of simply "AA".

For triangles ABC and ACD, we have the first pair of angles being A = A (angle A shows up twice each in the first slot). The second pair of congruent angles would be the right angles for triangle ABC and ACD, which are angles C and D respectively.

We can't use AAS because we don't know any information about the sides of the triangle.

7 0

2 years ago

Can someone help me plz

Lady bird [3.3K]

-4.25 because that is the only number out of those options between -4 and -5

8 0

3 years ago

Read 2 more answers

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

Can anyone help me with this and explain it too?

arlik [135]

Answer:

(6, 3)

Step-by-step explanation:

$Midpoint (x_{m},y_{m})=(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2})\\\\=(\frac{3 + 9}{2}, \frac{6 + 0}{2})\\\\=(\frac{12}{2}, \frac{6}{2})\\\\Midpoint of a line segment (x_M, y_M) = (6, 3)$

How to Calculate Midpoint?

The x-coordinate of the midpoint M of the segment $\overline{AB}$ is the arithmetic mean of the x-coordinates of the endpoints of the segment $\overline{AB}$ . Similarly, the y-coordinate of the midpoint M of the segment $\overline{AB}$ is the arithmetic mean of the y-coordinates of the endpoints of the segment $\overline{AB}$ .

7 0

3 years ago