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3 years ago

Which function produces an output of 5 for an input of -6?

Mathematics

1 answer:

kodGreya [7K]3 years ago

3 0

Answer:

The answer is D

Step-by-step explanation:

Since the input is -6, you input it into the equation. That would make it f(-6)=-(-6)-1 which would be equivalent to f(-6)=6-1 , as the negatives at the beginning of the equation cancel each other out and make it positive. Then you solve and get 5, which is your output. Hope I could help :)

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Which of the following choices represents the area of the figure below?

Agata [3.3K]

Answer:

x^2-16

Step-by-step explanation:

A= (b)(h)

A= (x+4)(x-4)

x^2-4^2

x^2-16

6 0

2 years ago

X/6=36/24 what value of x makes this proportion true

Stolb23 [73]

Cross multiply or just multiply both sides by 24
24x/6=36
4x=36
divide both sides by 4
x=9

4 0

3 years ago

Read 2 more answers

Complete the equation of variation where y varies inversely as x and y = 0.125 when x = 8.

patriot [66]

Complete the equation of variation where y varies inversely as x and y = 0.125
when x=8

3 0

3 years ago

a group of 31 friends gets together to play a sport. first people must be divided into teams. each team has to exactly 4 players

azamat

We have the following values:
Total people: 31
People per team: 4
The number of teams will then be:
N = (31) / (4)
N = 7.75
Round to the previous whole number.
N = 7
There are 7 teams.
Answer:
they can make about 7 teams.

3 0

3 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

4 0

3 years ago