All categories

4 years ago

A can holds 3 tennis balls. A box holds 4 cans. A case holds 6 boxes. How many tennis balls are in 10 cases?

Mathematics

2 answers:

Korolek [52]4 years ago

5 0

Answer:

720 balls

Step-by-step explanation:

Can = 3 balls

Box = 4 cans = 4*3balls = 12 balls

Case = 6 boxes = 6* 12balls = 72 balls

10 cases = 10 *72balls = 720 balls

myrzilka [38]4 years ago

4 0

Answer: 720

Step-by-step explanation:

If there are 10 cases and 6 boxes in a case. Multiply 10 by 6 to get 60.

Then if there are 4 cans in a box. Multiply 4 by 60 to get 240.

Finally, if there are 3 tennis balls per can. Multiply 240 by 3 to get 720.

There are 720 Tennis Balls in 10 Cases.

You might be interested in

Use the expression 34-12 divided by 2+7 place parentheses in the expression so that the value of the expression is 18

Ulleksa [173]

(34-12)÷2+7

22÷2+7

11+7

18

Hope this helps! If you want more details in case you don't understand, leave a comment, I'll reply as soon as possible! :)

8 0

3 years ago

Find sin ( a +b) if sin a + -20/29 where a is in the fourth quadrant and tan (b) = -15/8 where b is in the second quadrant

Rzqust [24]

Hope this helps! :))))

6 0

3 years ago

Please help me...asap

PolarNik [594]

All members of the sales department

4 0

2 years ago

Foil: (x+9)(3x-4)

QveST [7]

QUESTION 1

The given binomial is;

$(x+9)(3x-4)$

First terms are multiplied: $x\times 3x=3x^2$

Outside terms are multiplied: $x\times -4=-4x$

Inside terms are multiplied: $9\times 3x=27x$

Last terms are Multiplied: $9\times -4=-36$

This gives us;

$=3x^2-4x+27x-36$

$=3x^2+23x-36$

QUESTION 2

We want to factor

$6r^5s+4r^4s^2-8r^2s$

The HCF is $2r^2s$

We factor to get;

$2r^2s(3r^3+2r^2s-4)$

QUESTION 3;

$x^2+x-2=0$

Split the middle term;

$x^2+2x-x-2=0$

Factor

$x(x+2)-1(x+2)=0$

$(x+2)(x-1)=0$

$(x+2)=0,(x-1)=0$

$x=-2,x=1$

The solutions are;

$x=-2$ and $x=1$

These are the x-intercepts of the graph of the function

$f(x)=x^2+x-2$

3 0

4 years ago

Which of the following would produce a negative sum

KIM [24]

Answer:

um are there options????

5 0

3 years ago