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2 years ago

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Fabio and carlos play on a basketball team together. in the last game, fabio had 777 points less than 222 times as many points a

s carlos. fabio scored 313131 points in the game. write an equation to determine the number of points (c)(c)left parenthesis, c, right parenthesis carlos scored in the last game. find the number of points carlos scored in the last game. points

2 answers:

jasenka [17]2 years ago

5 0

For this case, the first thing we must do is define variables.

We have then:

f: number of points that fabio scored.

c: number of points that Carlos scored.

We now write the equation that models the problem:

$f = 2c - 7$

Then, for f = 31 we have:

$2c - 7 = 31$

From here, we clear the number of points:

$2c = 31 + 7$

$2c = 38$

$c = \frac{38}{2}$

$c = 19$

Answer:

The equation to find the number of carlos points is:

$2c - 7 = 31$

Then, Carlos scored:

$c = 19$

svlad2 [7]2 years ago

5 0

Answer:the correct equation is 2c-7=31

Step-by-step explanation:

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Starting with the paints, we have 15$ as our base value. Next, since we add $24.50 for each new canvas bought, 24.5*x=the amount of money for x canvases. Therefore, since 15 is our base value, we have 15+24.5*6=162

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Let L be the line with parametric equations x=2+t, y=1-t, z=1+3t.Let v=(1,2,0).Find vectors w1 and w2 such that v= w1 + w2, and

charle [14.2K]

Answer:

w1 = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

Step-by-step explanation:

Direction ratio of w1 = Direction ratio of L (because parallel) = K+(1, -1, 3)

Let <a,b,c> be direction ratio of w2.

Then, <a,b,c>. <1,-1,3> = 0

a-b+3c = 0

v = w1 + w2

(1, 2, 0) = k(1, -1, 3) + (a, b, c)

a + k = 1

b - K = 2

c - 3k = 0

Solving 4 equations, a = 12/11, b= 21/11, c = 3/11, k=-1/11

So, w1 = -1/11(1, -1 ,3) = (-1/11, 1/11, -3/11)

w2 = (12/11, 21/11, 3/11)

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The car you just bought keeps breaking down. The parts you need to buy cost $408.97. Your mechanic also charges $53 per hour for

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Find the GCF of 44j5k4 and 121j2k6.

BartSMP [9]

Answer:

D. $11j^2k^4$

Step-by-step explanation:

We are asked to find the GCF of $44j^5k^4\text{ and }121j^2k^6$ .

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of $j^5\text{ and }j^2$ .

Factors of $j^5$ are: $j*j*j*j*j$

Factors of $j^2$ are: $j*j$

We can see that greatest common factor of $j^5\text{ and }j^2$ is $j*j=j^2$ .

Now let us find GCF of $k^4\text{ and }k^6$ .

Factors of $k^4$ are: $k*k*k*k$

Factors of $k^6$ are: $k*k*k*k*k*k$

We can see that greatest common factor of $k^4\text{ and }k^6$ is $k*k*k*k=k^4$ .

Upon combining our all GCFs we will get,

$11j^2k^4$

Therefore, GCF of $44j^5k^4\text{ and }121j^2k^6$ is $11j^2k^4$ and option D is the correct choice.

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2 years ago

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The DiMonte Corporation invented a new type of sunglass lens. Their variable expenses are $12.66 per unit, and their fixed expen

Sladkaya [172]

Answer:

a)one lens: 111,212.66

15000 lenses: 301,100

b)y=111,200+12.66x

slope: 12.66

c)12.66 dollars per unit of lens produced

d) 20.07

e) 19.2

f)decreases

Step-by-step explanation:

a) Cost of producing one lens

fixed expenses+variable expenses=111,200+12.66=111,212.66

Cost of producing 15,000 lenses

fixed expenses+variable expenses=111,200+12.66(15,000)=301,100

b)Expense function:

let y be the expense cost

let x be the number of lenses produced

y=111,200+12.66x

Slope:

The function is a line equation: y=mx+b where m is the slope.

In this case b=111,200 and m=12.66. So the slope is 12.66

c) the cost increases 12.66 dollars per unit of lens produced

d) The cost of producing 15000 lenses is:

111,200+12.66(15,000)=301,100

If DiMonte produces 15000 lenses, the average cost per lens is:

301,100/15000=20.073≈20.07

e)The cost of producing 17000 lenses is:

111,200+12.66(17,000)=326,420

If DiMonte produces 17000 lenses, the average cost per lens is:

326,420/17,000=19.2

f)The average per lens decrease.

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