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3 years ago

11

jordan bought 12 tennis balls and annika bought 16. the tennis balls come in packages. each package has the same number. what is

the greatest possible number of tennis balls in each package

1 answer:

vova2212 [387]3 years ago

4 0

It would be 4 because it is the highest number that can go into 12 and 16.

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Suppose you have $16,000 to invest in three stocks, A, B, and C. Stock A is a low-risk stock that has expected returns of 4%. St

podryga [215]

To solve this problem, let us say that:

money invested in stock A = A

money invested in stock B = B

money invested in stock C = C

The given problem states that:

C = A * (1 / 4) = 0.25 A

B = A * (1 / 2) = 0.50 A

It was stated that we only have $16,000 to invest. Therefore:

A + B + C = 16,000

Substituting values of C and B in terms of A:

A + 0.50 A + 0.25 A = 16,000

1.75 A = 16,000

A = $9,142.86

So C and B is:

C = 0.25 (9142.86)
C = $2285.71

B = 0.50 (9142.86)

B = $4571.43

5 0

3 years ago

Read 2 more answers

What is -a^-2 if a=-5

Leviafan [203]

$-a^{-2}=-(-5)^{-2}=-\left(\dfrac{1}{-5}\right)^2=-\dfrac{1}{25}$

3 0

3 years ago

Country days scholarship rounds receive a gift of $135000. The money is invested in stock, bonds, and CDs. CDs pay 2.75% interes

Vsevolod [243]

Answer:

CDs — $10,000
bonds — $80,000
stocks — $45,000

Step-by-step explanation:

Let the variables c, b, s represent the dollar amounts invested in CDs, stocks, and bonds, respectively. Then the problem statement gives us 3 relations between these 3 variables:

c + b + s = 135000 . . . . . . . . . . . . . . . . . total invested

0.0275c +.045b +0.104s = 8555 . . . . . total income earned

-c + b = 70000 . . . . . . . . . . . . . . . . . . . . . 70,000 more was in bonds than CDs

Using the third equation to write an expression for b, we can substitute into the other two equations.

b = 70000 +c . . . . . . . . . . . . . . . . expression we can substitute for b

c + (70000 +c) +s = 135000 . . . . substitute for b in the first equation

2c +s = 65000 . . . . . . . . . . . . . . . . [eq4] simplify

.0275c +.045(70000 +c) +.104s = 8555 . . . . . substitute for b in 2nd eqn

.0725c +.104s = 5405 . . . . . . . . . . [eq5] simplify

Using [eq4], we can write an expression for s that can be substituted into [eq5].

s = 65000 -2c . . . . . . . expression we can substitute for s

0.0725c +0.104(65000 -2c) = 5405

-0.1355c = -1355 . . . . . . . . . . . . . . . . . . . . subtract 6760, simplify

c = 1355/.1355 = 10,000

s = 65000 -2×10000 = 45,000

b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

$\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]$

6 0

3 years ago

Find the exact value of tan 15 degrees

pentagon [3]

<span>tan(15) =
sin(15) / cos(15) =
sin(45 - 30) / cos(45 - 30) =
[ sin(45)cos(30) - sin(30)cos(45) ] / [ cos(45)cos(30) + sin(45)sin(30)]

Since sin(45) = cos(45) = √2/2, you can just factor that out from the top and bottom
[ cos(30) - sin(30) ] / [ cos(30) + sin(30)]
[ √3/2 - 1/2 ] / [ √3/2 + 1/2]
(√3 - 1) / (√3 + 1)
(√3 - 1)^2 / (√3+1)(√3 - 1)
(√3 - 1)^2 / (3 - 1)
(3 - 2√3 +1) / 2
2 - √3

There's also a formula for tan(a-b), but I couldn't remember it off hand.</span>

4 0

3 years ago

Charity is four times as old as her daughter. She is also six years younger than her husband. The sum of their ages ten years to

zubka84 [21]

Answer: Charity is 36 years old today.

Step-by-step explanation:

Let x = Present age of daughter.

Age of Charity = 4x

Age of her husband = 4x+6

After 10 years,

Age of daughter = x+10

Age of Charity = 4x+10

Age of her husband = 4x+6+10=4x+16

As per given,

$x+10+4x+10+4x+16=117$

$9x+36=117\\\\ 9x=117-36\\\\ 9x=81\\\\ x=9$

Present age of Charity = 4(9)=36 years

Hence, Charity is 36 years old today.

4 0

3 years ago