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4 years ago

14

Evaluate the expression when y = 5. 3 + y + 6 9 14 19 21

2 answers:

vekshin14 years ago

7 0

Yeerrrrreeeeeep ^^^^^%^

REY [17]4 years ago

6 0

Its prob b- 14 cause they all add up

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8-2x<4 solve inequality

Andreas93 [3]

Answer:

x > 2

x ∈ ( +2 ; + oo )

Step-by-step explanation:

8 - 2x < 4

8 - 4 < 2x

4 < 2x | : 2

2 < x

x > 2

x ∈ ( +2 ; + oo )

6 0

3 years ago

The diagram below shows several parking spots near the grocery store, formed with three parallel line segments and a transversal

Marina CMI [18]

Something I noticed about angle 4, is that it shares a straight line with angle 3. (above it is a straight line)

A straight line equals 180 degrees.

So, angle 4 + angle 3= 180 degrees.

Angle 3 is congruent to angle 7.

180-132= Angle 7

180-132= 48

48 degrees is your answer.

I hope this helps!
~kaikers

8 0

4 years ago

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Can someone help me understand this? Because I really don’t understand this at all

Otrada [13]

Answer: I believe it is A OR B

Step-by-step explanation: BECAUSE IF A⇒B AND B⇒C

THEN ¬A⇒C I THINK ITS A THO

OR ¬A⇒¬C

7 0

3 years ago

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

The answer is 60, but I don’t know how to fill in the chart. Idk what I’m comparing.

FromTheMoon [43]

Answer:

you have to write the way you did the problem

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers