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4 years ago

5

To convert degrees Fahrenheit (F) into degrees Celsius (C) use the formula 2003-05-04-00-00_files/i0150000.jpg. Rewrite the equa

tion to convert degrees Celsius to degrees Fahrenheit.

1 answer:

bulgar [2K]4 years ago

5 0

Here's the formula for both conversions:

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In each one section there are 27 student and there are 17 section how many students are in total

stiks02 [169]

Answer:

459 students.

Explanation:

Since you are given the number of how many students are in each section, simply multiply to find the total answer.

27 x 17 = 459

There are 459 students.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!!!!!

Ne4ueva [31]

Answer:

The first 5 terms are;

-2, 2, 13,38 and 91

Step-by-step explanation:

Here, we want to write the first 5 terms of the sequence.

We already have the first term as 1

Now, we need the 2nd term

Putting two in place of n, we have ;

2f(1) + 3n

= 2(-2) + 3(2) = -4 + 6 = 2

For the 3rd term, put 3 in place of n

2f(2) + 3(3)

sine f(2) = 2, we have

2(2) + 9 = 4+ 9 = 13

For the fourth term, put 4 in place of n, we have

2f(3) + 3(4)

since f(3) = 13

we have; 2(13) + 12 = 26 + 12 = 38

For the 5th term, put 5 in place of n, we have

2f(4) + 3(5)

since f(4) = 38, we have

2(38) + 15 = 76 + 15 = 91

6 0

3 years ago

A scale drawing of a school playground in the shape of a rectangle is shown. What is the area in square meters of the playground

Ilya [14]

Answer:

b

Step-by-step explanation:

8×19= 152

1 : 4.5

152×4.5=684

3 0

3 years ago

******ANSWER ASAP PLEASE******

Lemur [1.5K]

(1/4)^-2 - (5^0 x 2) x 1^-1 =

(4/1)^2 - (1 x 2) x 1 = 16-2 = 14

If you raise something to the power of -2, swap numerator and denominator and remove the minus.

So (1/4)^-2 = 4^2 = 16

Also 1^-1 is just 1, not -1.

8 0

3 years ago

Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x

tester [92]

Answer:

The solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

Step-by-step explanation:

We will need the double angle identity $\sin(2x)=2\sin(x)\cos(x)$ .

Let's begin:

$2\sin(2x)=\sin(x)$

Use double angle identity mentioned on left hand side:

$2\cdot 2\sin(x)\cos(x)=\sin(x)$

Simplify a little bit on left side:

$4\sin(x)\cos(x)=\sin(x)$

Subtract $\sin(x)$ on both sides:

$4\sin(x)\cos(x)-\sin(x)=0$

Factor left hand side:

$\sin(x)[4\cos(x)-1]=0$

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

$\sin(x)=0 \text{ or } 4\cos(x)-1=0$

The first is easy what angles $\theta$ are $y$ -coordinates on the unit circle 0. That happens at $0$ and $\pi$ on the given range of $x$ (this $x$ is not be confused with the $x$ -coordinate).

Now let's look at the second equation:

$4\cos(x)-1=0$

Isolate $\cos(x)$ .

Add 1 on both sides:

$4 \cos(x)=1$

Divide both sides by 4:

$\cos(x)=\frac{1}{4}$

This is not as easy as finding on the unit circle.

We know $\arccos( )$ will render us a value between $0$ and $2\pi$ .

So one solution on the given interval for x is $x=\cos^{-1}(\frac{1}{4})$ .

We know cosine function is even.

So an equivalent equation is:

$\cos(-x)=\frac{1}{4}$

Apply $\cos^{-1}$ to both sides:

$-x=\cos^{-1}(\frac{1}{4})$

Multiply both sides by -1:

$x=-\cos^{-1}(\frac{1}{4})$

This going to be negative in the 4th quadrant but if we wrap around the unit circle, $2\pi$ , we will get an answer between $0$ and $2\pi$ .

So the solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

5 0

3 years ago