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3 years ago

5

HELPPPPPPPW PLSSSSBRBEHD

2 answers:

Strike441 [17]3 years ago

7 0

Answer:

b. 10

Step-by-step explanation:

3 (x 2) = 6

5 (x 2) = 10

dezoksy [38]3 years ago

3 0

Answer:

Helen scores <u>10</u> points.

Step-by-step explanation:

We are given that for every <u>3 points</u> Glenn scores, Helen scores <u>5 points</u>. Glenn now scores <u>6 points</u> and Helen's score is unknown. We can write this as a ratio.

Glenn - 3 : 5 - Helen

Glenn - 6 : ? - Helen

We can divide 6 by 3 to get the rate of change, which is 2. Then multiply 5 and 2 and we get 10! Helen scores 10 points for every 6 points Glenn scores. I hope this helps! :D

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HELP ASAP PLZZZ!!! I would be Soo grateful

Dmitry [639]

Answer:

31%

Step-by-step explanation:

3/10 is the same as 30%

If 30% of the students liked English and 61% of the students like math, the difference can be found by using subtraction.

61% - 30% = 31%

5 0

3 years ago

Find the average rate of change of the function over the given interval.

Minchanka [31]

Answer:

Average rate of change = 20.2

Rate of change at the left endpoint : f' (5) = 4t = 20

Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4

Step-by-step explanation:

The average rate of change of the function

F(t) = 2t^2 - 1 , [ 5,5.1 ]

solution

$\frac{f(5.1)-f(5)}{5.1 - 5}$ = $\frac{2*(5.1)^2-1-(2*5^2-1)}{0.1}$ = [ 2 ( 26.01 - 25 ) / 0.1 ]

= 2.02 / 0.1 = 20.2

Rate of change at the left endpoint : f' (5) = 4t = 20

Rate of change at the right endpoint : f' (5.1) = 4*5.1 = 20.4

6 0

4 years ago

Which equation is graphed here?

Oxana [17]

The equation graphed is

y = -3x + 3

5 0

3 years ago

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

4 0

3 years ago

Point F is reflected over the y-axis to create F’. Use an ordered pair to name the location of F’, and determine the distance be

bagirrra123 [75]

Answer:

the location of F' is (6,3)

6 0

3 years ago