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3 years ago

Write 12 more than r as an expression

Mathematics

1 answer:

vaieri [72.5K]3 years ago

7 0

Answer:

12+r

Step-by-step explanation:

"more than" means Addition.

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What is the Simplified form of √46/192

Lorico [155]

Answer:

(√138)/24

Step-by-step explanation:

$\displaystyle\sqrt{\frac{46}{192}}=\sqrt{\frac{46\cdot 3}{192\cdot 3}}=\sqrt{\frac{138}{24^2}}=\frac{\sqrt{138}}{24}$

8 0

3 years ago

If f(x)=3x+2 what is the equation for f^-1(x)

Elden [556K]

Answer:

f^-1(x) = (x -2)/3

Step-by-step explanation:

To find the inverse function, swap x and y, then solve for y.

y = 3x +2 . . . . given function

x = 3y +2 . . . . inverse relation

x -2 = 3y . . . . subtract 2

(x -2)/3 = y . . . divide by 3

f^-1(x) = (x -2)/3 . . . . in functional form

4 0

3 years ago

50 POINTS AND BRAINLEST(if its right)

zhenek [66]

Answer:

X=-0.6

Step-by-step explanation:

Hope it help

3 0

3 years ago

Read 2 more answers

Set up and solve a quadratic equation to work out the value of x

galben [10]

X= 24 simple calculations are there

4 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ subtract 16 from both sides

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$

5 0

3 years ago