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3 years ago

7

6. At a game, there were 24 adults and 10 children. Keeping the same ratio,

1 answer:

irina1246 [14]3 years ago

6 0

Answer:

The total number of adults will be: 120 adults

Step-by-step explanation:

Given that there were 24 adults and 10 children in a game.

Thus, adults to children ration = 24:10 or 24/10

If there were 50 children, how many adults would be at the game?

Let 'x' be the number of adults when there 50 children.

so the ration of adults 'x' to 50 childern = x: 50 or x / 50

As the ratio is the same.

Thus,

24 : 10 = x : 50

24/10 = x/50

switching sides

x/50 = 24/10

x = 24/10 × 50

x = 120 adults

Therefore, the total number of adults will be: 120 adults

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A small plane flew 808 miles in 4 hours with the wind. Then on the return trip flying against the wind it travels only 488 miles

Blababa [14]

Answer:

Step-by-step explanation:

let wind velocity=x

speed of plane=y

(x+y)*4=808

x+y=808/4=202 ...(1)

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y-x=122 ...(2)

add (1) and (2)

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y=324/2=162

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2 years ago

What is t+1/5=t+7/8? What is the value of t? There should be 2 values and the slashes are fractions

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There is NO solution.

Cancel t on both sides (1/5 = 7/8).

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3 years ago

Priya has completed 9 eczema questions.THis is 60% of the questions on the exam? how many questions are on the exam?

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3 years ago

Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

$x^{2}+x-56=0$

$x^{2}+8x-7x-56=0$

$(x+8)-7(x+8)=0$

$(x+8)(x-7)=0$

$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

(b) Since $x=-8,7$ are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

$f(7)=2(7)^3+3(7)^2-336(7)$

$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

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3 years ago