What is c in this ixl

Answer:

  period

Step-by-step explanation:

The attached graph shows a period of the function. Its amplitude is 1/4 (vertical distance from centerline to peak). Its period is 4π (horizontal distance from peak to peak). Its frequency is the reciprocal of the period, so would be 1/(4π).

The value 4π is the period of the function.

Answer:

x= 12

Step-by-step explanation:

Hope u understood it

"PLEASE MARK ME THE BRAINLIEST"

Answer:

it is A Scalene Triangle

5 quarts I think 4 cups per quart has 5 quarts and 2 cups I think

Answer:

1 quart = 4 cups

8 × 4 = 32

32 - 6 = 26

26 - 18 = 8

he has 2 quarts left

Answer:

a

Step-by-step explanation:

7(g-9)+12=26

7(g-9)+12-12=26-12

7(g-9)=14

7(g-9)/7=14/7

g-9=2

g-9+9=2+9

g=11

Answer:

Step-by-step explanation:

26=7(g-9)+12

(take 12 to 26)

so it will be 7(g-9)= 26-12

=> 7(g-9)= 14

(then take 7 to 14)

=> g-9= 14/2

=> g-9= 2

(then take 9 to 2)

g= 2 + 9

hope it is properly explained

STAY SAFE, HAVE A NICE DAY

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

As we know ~

Area of the circle is :

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 4.4\div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.2 \: mm[/tex]

Now find the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.2)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {4.84}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 3.7 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1.85 \: \: cm[/tex]

Bow, calculate the Area ~

[tex]\qquad \sf  \dashrightarrow \: \pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{} [/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 68.89[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx216.31 \: \: cm {}^{2} [/tex]

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[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 5.8 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 2.9 \: \: yd[/tex]

now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(2.9)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 8.41 [/tex]

[tex] \qquad \sf  \dashrightarrow \:area \approx26.41 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:r = d \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 1 \div 2[/tex]

[tex]\qquad \sf  \dashrightarrow \:r = 0.5 \: \: yd[/tex]

Now, let's calculate area ~

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 0.25[/tex]

[tex]\qquad \sf  \dashrightarrow \:area \approx0.785 \: \: yd {}^{2} [/tex]

・ ．━━━━━━━†━━━━━━━━━．・

[tex]\qquad \sf  \dashrightarrow \:\pi {r}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times {(8)}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:3.14 \times 64[/tex]

[tex]\qquad \sf  \dashrightarrow \:area = 200.96 \: \: yd {}^{2} [/tex]

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Answer:

4 boxes

Step-by-step explanation:

If she has 4 boxes that hold 5 cookies each, that will total 20 cookies.

4×5=20

20÷5=4

Step-by-step explanation:

inscribed angles subtended by the same arc are equal.

the central angle of a circle is twice any inscribed angle subtended by the same arc.

the first statement tells us that the 53° angle as well as y stay the same size no matter where on their arcs (between the 2 points connected to O) they would be. so, we don't need to bother with any line lengths.

the 2nd statement tells us that x = 2×53 = 106°. the 53° and x angles refer to the short arc on the right of the 2 points connected to O.

and y and x refer to the larger arc on the left of the 2 line connected to O. that means according to the second statement : 360-x (the big angle around O) = 2y

so,

360 - 106 = 2y

254 = 2y

y = 127°

Answer:

2929 1/3 [tex]\pi[/tex] in^3

Step-by-step explanation:

The equation for the volume of a sphere is [tex]\frac{4}{3} \pi r^3[/tex].

Insert 13 for r.  Then, simplify.

[tex]\qquad\qquad\huge\underline{\boxed{\sf Answer}}[/tex]

Let's solve ~ ☂

⧠ Here, I'm using middle term split method :

[tex] \qquad\rightarrowtail \sf {x}^{2} + 6x + 8[/tex]

[tex] \qquad\rightarrowtail \sf {x}^{2} + 4x + 2x+ 8[/tex]

[tex] \qquad\rightarrowtail \sf {x(x}^{} + 4) + 2(x+ 4)[/tex]

[tex] \qquad\rightarrowtail \sf (x + 2)(x+ 4)[/tex]

[tex]\\ \rm\rightarrowtail x^2+6x+8[/tex]

[tex]\\ \rm\rightarrowtail x^2+4x+2x+8[/tex]

[tex]\\ \rm\rightarrowtail x(x+4)+2(x+4)[/tex]

[tex]\\ \rm\rightarrowtail (x+2)(x+4)[/tex]

Done!

Answer:

56

Step-by-step explanation:

Answer:

5.6

Step-by-step explanation:

4/5 = .8

so ? ÷ 7 = .8|
7 x .8 = 5.6
x= 5.6

x = 5.6

hope it helps...!!!!

Answer:

As per Provided Information

Height of cone h is 3.25 cm

Diameter of cone is 2.2 cm

Radius = 2.2/2

Radius = 1.1 cm

We have been asked to determine the volume of the cone .

Using Formulae

[tex] \boxed{\bf \:Volume_{(Cone)} = \cfrac{1}{3} \pi {r}^{2}h}[/tex]

Substituting the given value and let's solve it

[tex]\sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times {1.1}^{2} \times 3.25 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times 1.21 \times 3.25 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 3.14 \times 3.9325 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{1}{3} \times 12.34805 \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = \cfrac{12.34805}{3} \\ \\ \\ \sf\longrightarrow \: Volume_{(Cone)} = 4.11 {cm}^{3} [/tex]

Therefore,

Answer:

30.37

Step-by-step explanation:

Answer:

[tex]164 \div 5.40 ={\ 30.3704{}}[/tex]

The answer is 30.3704

Answer:

13 C 14 D 15 A 16 B

Step-by-step explanation:

Sorry thats all i know hope you found this helpful

Answer:

2.7

Step-by-step explanation:

Answer:

you do not have any numbers, only letters so I cannot answer your question. sorry...

Step-by-step explanation:

Select the correct answer.

Which number is an irrational number?

A.

B.

C.

D.

Answer:

[tex]P(A < 84)=\frac{4}{7}[/tex]

Step-by-step explanation:

[tex]Total\ numbers:\ 7\\Numbers\ less\ than\ 84:\ 4\\\\P(A < 84)=\frac{4}{7}[/tex]

Answer:

f(-6)= -41 923

Step-by-step explanation:

f(-6)= -( -6)⁶ +5(-6)⁴+8(-6)³+2(-6) -7 you substitute x with -6 the you use a calculator which will then give you - 41923

Answer:

71.428571%

Step-by-step explanation:

You divide 5 by 7

Answer:

Step-by-step explanation:

you would multiply the fractions

then you would add the hole number

Answer:

2 14/15

Step-by-step explanation:

Answer:

C. 3000

Step-by-step explanation:

i took the quiz!

also 60 and 3000 line up

Answer:

C 3,000

Step-by-step explanation: It's right on Edge

Answer:

Pretty sure it's 2:30 P.M.

Step-by-step explanation:

If her track practice started at 12:00 P.M., and and ran relays for one hour, it would already be 1:00 P.M.

Then she practiced the long jump for an hour and thirty minutes. Add another hour to that and it would be 2:00 P.M. Add thirty minutes, and it would be 2:30 P.M.

Hopefully that makes sense to you

Answer:

5π/6

Step-by-step explanation:

Given :

Solving :

Answer:

5π Over 6

Step-by-step explanation:

You can write  7π6 as (π+π6)Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative. Hence, cos(π+π6)=−cos(π6)coming back to the question cos−1[cos(7π6)]=cos−1[−cos(π6)]=π−cos−1[cos(π6)]=π−π6=5π6

The calculated areas of both are the same instead of using the different formulas that are 43. 5 squared unit.

The area of the polygon is the product of half of the apothem of the triangle and perimeter.

The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]

The side of the triangle is 10 units.

Apothem of an equilateral triangle

[tex]s = \frac{\sqrt{3} a}{6}[/tex]

s= [tex]\sqrt{3} \times 10/6[/tex]

s = 2.88

Here, a is the side of the triangle.

Thus the apothem of the triangle is 2.9 units.

The perimeter of the equilateral triangle is equal to the three times the sides.

Perimeter of the equilateral triangle ,

P = 3(10)

P = 30

Therefore, the perimeter of the equilateral triangle is 30 units.

The area of the equilateral triangle-

The area of the polygon = [tex]\frac{1}{2} \times h \times p[/tex]

A = [tex]1/2 \times 2.9 \times 30[/tex]

A = 43.5

Thus the area of the equilateral triangle is 43.5 squared units.

Therefore, The calculated areas of both are the same instead of using the different formulas.

Learn more about the apothem;

brainly.com/question/10580427

Answer:

(A) 2.9

(C)30

(A)1/2(2.9)(30)

(D)The same despite using different formulas

Step-by-step explanation:

I did it

Answer:

-13

Step-by-step explanation:

Answer:

[tex]\displaystyle 11,3[/tex]

Step-by-step explanation:

Use the Law of Cosines to find the length of the third edge:

Solving for Angles

[tex]\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos\angle{C} \\ \frac{a^2 - b^2 + c^2}{2ac} = cos\angle{B} \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos\angle{A}[/tex]

Use [tex]\displaystyle cos^{-1}[/tex]towards the end or you will throw your result off!

Solving for Edges

[tex]\displaystyle b^2 + a^2 - 2ba\:cos\angle{C} = c^2 \\ c^2 + a^2 - 2ca\:cos\angle{B} = b^2 \\ c^2 + b^2 - 2cb\:cos\angle{A} = a^2[/tex]

Take the square root of the final result or it will be thrown off!

Let us get to wourk:

[tex]\displaystyle 10^2 + 7,9^2 - 2[10][7,9]\:cos\:77,5 = a^2 \hookrightarrow 100 + 62,41 - 158\:cos\:77,5 = a^2 \hookrightarrow \sqrt{128,212541} = \sqrt{a^2}; 11,323097677... \\ \\ \boxed{11,3 \approx a}[/tex]

I am joyous to assist you at any time.