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damaskus [11]
3 years ago
13

13. Solve for x so that f(x) = 5​

Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

x=\frac{5}{f}

Step-by-step explanation:

You might be interested in
Please help me thanks
jasenka [17]

Answer:

126

Step-by-step explanation: just multiply 9 and 13

6 0
3 years ago
Read 2 more answers
6. The price of a pack of toilet rolls was increased by 20%.
Ymorist [56]
Ok thanks for the points
5 0
3 years ago
Can someone help me with number five ???
exis [7]
Hey there! This is gonna actually be surprisingly simple even though it looks confusing!

Since the scale is perfectly balanced (it's completely flat because the weights are equal) you can use an = sign.

There are 3 cubes which weigh 9 ounces so 3 times 9 is 27 so on the left side you can put down a 27. Then there are 7 circles so so 7x can represent this. So that belongs on the left side as well SO for the left sides we can write this
27 + 7x = _______

Now for the right side. There are 5 cubes so 5 times 9 ounces is 45 so that belongs on the left. And then there is one cube so we can just right x. SO for the right side (combined with the other parts we know looks like this.
27 + 7x = 45 + x

NOW for the solving!
<span>27 + 7x = 45 + x
</span>subtract x from both sides
27 + 6x = 45
subtract 27 from both sides
6x = 18
then divide 6 from both sides
x = 3

So the circles are 3 ounces!!!!

I hope that helps! And feel free to ask me any questions!!

- mathwizzard3
5 0
3 years ago
pleaseee help how do i write an equation in slope-intercept form for the line with slope 1/3 and y-intercept -3?
Montano1993 [528]

Answer:

y=1/3x -3

Step-by-step explanation:

Slope intercept form  is y=mx+b, where m is the <u>slope </u>and b is the <u>y-intercept, </u>or the y-value at which the line intersects the y-axis.

Your slope is 1/3, and your y-int. is -3, so substitute those values in:

y=<u>1/3</u>(x) <u>- 3</u>

8 0
3 years ago
Use a half-angle identity to find the exact value
Tatiana [17]

Given:

\cos 15^{\circ}

To find:

The exact value of cos 15°.

Solution:

$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}

Using half-angle identity:

$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}

$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}

Using the trigonometric identity: \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}

            $=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}

Let us first solve the fraction in the numerator.

            $=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}

Using fraction rule: \frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}

            $=\sqrt{\frac {2+\sqrt{3}}{4}}

Apply radical rule: \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}

           $=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}

Using \sqrt{4} =2:

           $=\frac{\sqrt{2+\sqrt{3}}}{2}

$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}

5 0
3 years ago
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