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3 years ago

Find the circumference of the circle. round your answer to the nearest centimeter.

Mathematics

1 answer:

Julli [10]3 years ago

7 0

What is the radius of the circle?

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1O less than the quotient of a number c and 13

Verizon [17]

Answer:

c/13 - 10

Step-by-step explanation:

5 0

1 year ago

What is the measure of the missing angle?

irinina [24]

125+42= 167

180-167= 13 degrees

Triangles is always equaled to 180 degrees.

Answer: choice 3

I am pretty sure this is the answer

6 0

2 years ago

Read 2 more answers

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a

Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago

Easy Area problem <br><br> Need it fast

Gekata [30.6K]

Answer:

just subtract the area of the rectangle from the area of the carpet

fine I'll do it

Area of a rectangle= lb

Area of a square= a^2

Area of the remaining area= lb - a^2

= (12*4) - (2*2)

= 48-4

= 44 ft^2

5 0

3 years ago

In 3-4 sentences, describe how you would find a line parallel to the line 2x + 5y = 15 that goes through the point (-10, 1). Be

larisa86 [58]

the parallel line is 2x+5y+15=0.

Step-by-step explanation:

ok I hope it will work

soo,

Solution

given,

given parallel line 2x+5y=15

which goes through the point (-10,1)

now,

let 2x+5y=15 be equation no.1

then the line which is parallel to the equation 1st

2 x+5y+k = 0 let it be equation no.2

now the equation no.2 passes through the point (-10,1)

or, 2x+5y+k =0

or, 2*-10+5*1+k= 0

or, -20+5+k= 0

or, -15+k= 0

or, k= 15

putting the value of k in equation no.2 we get,

or, 2x+5y+k=0

or, 2x+5y+15=0

which is a required line.

6 0

11 months ago