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ludmilkaskok [199]

3 years ago

11

What is the radius of a circle with with circumference 94.2mm?

2 answers:

irakobra [83]3 years ago

7 0

Answer:

14.99mm

Step-by-step explanation:

MAXImum [283]3 years ago

4 0

Answer: radius is 14. 99mm rounded up to 15.

Explanation:

C = 2πr

94.2 = 2π (3.14)r

94.2 = 6.28r

14.99 = r That's the radius!!

Hope this helps! :)

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dangina [55]

Answer:

$(x-3)^2 =21$

Step-by-step explanation:

$~~~~~~~x^2-6x-12=0\\\\\implies x^2 -6x +9-21=0 \\\\\implies x^2 -2\cdot x \cdot 3 +3^2=21\\\\\implies (x-3)^2 =21$

3 0

3 years ago

How many times does the graph of 4x=32-x^2 cross the x axis

qaws [65]

The number of times a graph crosses the x axis is the same as asking how many zeroes has the function represented by that graph.
The number of zeroes is given by the power of the independent variable, in this case x.
Given that the maximum power of x is 2, then the function has 2 zeroes and the graph crosses the x axis 2 times

6 0

3 years ago

Read 2 more answers

Evaluate this equation: (4^-3) (4^6)

marishachu [46]

The answer is -1/64, i hope its helps

4 0

3 years ago

Read 2 more answers

PLZ HELP WITH THIS QUESTION PLZ CORRECT ANSWERS!!! LOVE YOU THNX

iogann1982 [59]

Answer:

1) The value of x= 0.0642

2) $sin\theta=\frac{P}{H}=\frac{24}{25}$

$cos\theta=\frac{B}{H}=\frac{7}{25}$

$tan\theta=\frac{P}{B}=\frac{24}{7}$

$cosec\theta=\frac{H}{P}=\frac{25}{24}$

$cot\theta=\frac{B}{P}=\frac{7}{25}$

Step-by-step explanation:

1) Given : A right angle triangle with sides x and 10 and angle is 50°

To find : Value of x

Solution : Since, it is a right angle triangle

Therefore, we apply $cos\theta=\frac{B}{H}$

Where Base = x, Hypotenuse= 10, angle = 50°

$cos(50)=\frac{x}{10}$

$0.642=\frac{x}{10}$

$0.0642=x$

The value of x= 0.0642

2) Given : $sec\theta=\frac{25}{7}$

To find : Value of $sin\theta,cos\theta,tan\theta,cosec\theta,cot\theta$

Solution : We have given $sec\theta=\frac{25}{7}$

$sec\theta=\frac{H}{B}$

so, B= 7 , H=25

By Pythagoras theorem,

$H^2=P^2+B^2$

$25^2=P^2+7^2$

$625-49=P^2$

$\sqrt{576}=P^2$

$P=24$

Now, P=24, B=7, H=25

$sin\theta=\frac{P}{H}=\frac{24}{25}$

$cos\theta=\frac{B}{H}=\frac{7}{25}$

$tan\theta=\frac{P}{B}=\frac{24}{7}$

$cosec\theta=\frac{H}{P}=\frac{25}{24}$

$cot\theta=\frac{B}{P}=\frac{7}{25}$

5 0

3 years ago

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago