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4 years ago

6

A basketball player averages 21 points per game for the first 8 games of the season. if he is held scoreless in the next two gam

es, how many average points per game must he score in his final 6 games in order to complete the season with an overall average of 21 points per game?

1 answer:

Ne4ueva [31]4 years ago

8 0

Suppose the average the average points that the player must have in the six remaining games in order to finish with 24 game be x.
The total number of games in a season is :
(8+2+6)=14
total points for the 10 games will be:
a/10=21
a=210
thus the total score for the remaining 6 games should be:
(x+210)/14=21
x=84
This implies that the average score per game should be 84/6=14

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A car was sold at $34000. That price was 85℅ of the orginal price. What was the original price?

Art [367]

Hey there! The answer is $40000

<span>A car was sold at $34000. That price was 85℅ of the orginal price.
When we know 85% of the original price we can find 1% of this price by dividing by 85.

1% of the original price = $34000 / 85 = $400
</span>When we know 1% of the original price we can find 100% of this price by multiplying by 100.

100% of the original price = original price = $400 * 100 = $40000
The answer is $40000

~ Hope this helps you!

3 0

4 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

<u>case A)</u> $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> <u>the term is not a perfect cube</u>

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

<u>case B)</u> $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> <u>the term is not a perfect cube</u>

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

<u>case C)</u> $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> <u>the term is not a perfect cube</u>

$a^{22}$ ------> <u>the term is not a perfect cube</u>

$b^{16}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

<u>case D)</u> $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

<u>case E)</u> $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

Read 2 more answers

Will give brainliest to whoever figures this out :(

IrinaK [193]

Step-by-step explanation:

I think it's 8.9932×10⁶

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3 years ago

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Write a story problem to represent 3÷5=3/5

DIA [1.3K]

3÷5=3/5
3/5=0,6
3÷5=0,6

7 0

3 years ago

Solve for xxx. Write the smaller solution first, and the larger solution second.

____ [38]

Answer:

$x=-5\text{ or }x=2$

Step-by-step explanation:

We have been given an equation $x^2+3x-10=0$ . We are asked to find the solutions of our given equation.

We will use splitting the middle term method to solve quadratic equation.

We will split middle term of our given equation into parts such that whose product is $-10$ and whose sum is 3.

We know that such two numbers are $5\text{ and }-2$ .

$x^2+5x-2x-10=0$

$(x^2+5x)+(-2x-10)=0$

$x(x+5)-2(x+5)=0$

$(x+5)(x-2)=0$

Using zero product property, we will get:

$(x+5)=0\text{ or }(x-2)=0$

$x+5=0\text{ or }x-2=0$

$x=-5\text{ or }x=2$

Therefore, the solutions for our given equation are $x=-5\text{ or }x=2$ .

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3 years ago

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