Karim says the ratio of the distance he ran to the distance Shawn ran 4:3 give three possibilities for the distance each could have run

Answers

Answer 1

Answer: Karim=4,8km,12km,16km Shawn=3,6km,9km,12km

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What does r equal in 4r+5=3r

80 POINTSSolve for x. Use the completeing the square method. x^2+8x=7

Answers

Answer:

x = -4 ±√23.

Step-by-step explanation:

x^2 + 8x = 7

(x + 4)^2 - 16 = 7

(x + 4)^2 = 23

x + 4 = ±√23

x = -4 ±√23.

Answer: The correct answer is C.

Step-by-step explanation:

⇒ $x^(2) +8x+16$

⇒ which is also, $(X+4x^(2) ) = 23$

⇒ So, $x + 4$ is a square root of 23.

⇒ Then, it could be positive or negative one:

So that's why the answer would be C.

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Plz help #s 13....14....15.....16....17......18

Answers

#14 is B, #16 is A, #17 is B #18 is C. I can't read #15 or #13. If you need to show work i will say in a differ comment.

Tennis balls with a diameter of 2.5 in. are packaged 3 to a can. The can is a cylinder. Find the volume of the space in the can that is not occupied by tennis balls. Assume that the balls touch the sides, top, and bottom of the can. Round your answer to the nearest hundredth. What is the volume not occupied by balls?

Answers

a) First we calculate the volume of each ball:

$\boxed{V=(4\pi r^3)/(3)}\n \n \boxed{V=(4*(3,14)*(1.25)^3)/(3)=8,177 \ square \ inch}$

b) Now we calculate the volume of tree balls:

$\boxed{V_T=3*8,177=24.531 \ square \ inch}$

c) Now we calculate the volume of cylinder:

c1) Base area:

$\boxed{A_b=\pi r^2}\n \n \boxed{A_b=3,14*(1,25)^2=4.906 \ square \ inch}$

c2) Cylinder height

$\boxed{h=3*(2.5)=7.5 \ inch}$

c3) Cylinder Volume:

$V_C=A_b*h\n \n \boxed{V_C=4.906*7.5=36.795 \ cubic \ inch}$

d) Finally we calculate the internal space:

$\boxed{\boxed{s=36.795-24.531= 12,264\ cubic \ inch}}$

$d=2.5 \ in \n \n r=(d)/(2)=(2.5)/(2)=1.25 \ in \n \n \pi=3.14 \n \nVolume \ of \ a \ Cylinder : \n \nV = \pi r^2 h$

$h=3\cdot d = 3*2.5 = 7.5 \ in \n \nV_(c) = 3.14\cdot (1.25)^2\cdot 7.5 = 23.55 * 1.5625=36.80 \ in^3$

$The \ volume \ of \ tree \ balls : \n \n V_(3b)= 3 \cdot (4)/(3) \pi r^3 = 4\pi r^3 \n \nV_(3b)=4\cdot 3.14\cdot (1.25)^3= 12.56\cdot 1.9531 \approx 24.53 \ in^3 \n \nV_(c)-V_(3b)=36.80-24.53 =12.27 \ in^3 \n \n Answer : \ The \ volume \ not \ occupied \ by \ balls \ it \ 12.27 \ in^3$

I made up of six congruent squares attached by their side of form a t. what net am i