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Gelneren [198K]

2 years ago

8

there are 148 juniors and 272 seniors attending prom this year. if there must be three chaperones per 60 students, how many chap

erones are needed

1 answer:

valkas [14]2 years ago

3 0

21 chaperones are needed for 420 students.

Step-by-step explanation:

No. of juniors = 148

No. of seniors = 272

Total = 148+272 = 420 students

3 chaperones = 60 students

1 chaperone = $\frac{60}{3}=20\ students$

Ratio of chaperon to students = 1:20

Let,

x be the number of chaperons for 420 students.

Ration of chaperones to students = x:420

Using proportion

Ratio of chaperones to students :: Ratio of chaperons to students

$1:20::x:420$

Product of mean = Product of extreme

$20*x=420*1\\20x=420\\$

Dividing both sides by 20

$\frac{20x}{20}=\frac{420}{20}\\x=21$

21 chaperones are needed for 420 students.

Keywords: ratio, proportion

Learn more about ratios at:

#LearnwithBrainly

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What is -4√12+√75 in the simplest radical form?

nekit [7.7K]

Answer: -3 $\sqrt{3}$

Step-by-step explanation:

-4 $\sqrt{12}$ + $\sqrt{75}$

$\sqrt{12}$ can be written as $\sqrt{4}$ x $\sqrt{3}$

which is the same as 2 $\sqrt{3}$

Also , $\sqrt{75}$ can be written as $\sqrt{25}$ x $\sqrt{3}$ , which is the same as 5 $\sqrt{3}$

Substituting into the main question , we have

-4 (2 $\sqrt{3}$ ) + 5 $\sqrt{3}$

= -8 $\sqrt{3}$ + 5 $\sqrt{3}$

= -3 $\sqrt{3}$

3 0

2 years ago

Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3

Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

$f(x)=x^4-2ax^3+ax^2- x+k$

and

$f(-1)=3$

Answer:

$k = -2$ and $a=1$

Step-by-step explanation:

Given

$f(x)=x^4-2ax^3+ax^2- x+k$

$Factor:\ x - 4$

$f(-1)=3$

Required

Find a and k

For $f(-1)=3$

Substitute -1 for x

$f(x)=x^4-2ax^3+ax^2- x+k$

$f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k$

$f(-1) = 1 - 2a *-1 + a*1 +1 + k$

$f(-1) = 1 +2a + a +1 + k$

$f(-1) = 2 +3a + k$

Substitute 3 for f(-1)

$3 = 2 +3a + k$

Collect Like Terms

$3 - 2 = 3a + k$

$1 = 3a + k$

Also:

If $x - 2$ is a factor, then

$f(2) = 0$

Substitute 2 for x and 0 for f(x)

$f(x)=x^4-2ax^3+ax^2- x+k$

$0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k$

$0 = 16 - 2a * 8 + a * 4 - 2 + k$

$0 = 16 - 16a + 4a - 2 + k$

$0 = 16 - 12a - 2 + k$

Collect Like Terms

$2 - 16 = - 12a + k$

$-14 = k - 12a$

$k = 12a - 14$

Substitute 12a - 14 for k in $1 = 3a + k$

$1 = 3a + 12a - 14$

$1 = 15a - 14$

Collect Like Terms

$15a = 1 + 14$

$15a = 15$

Solve for a

$a = \frac{15}{15}$

$a=1$

Substitute 1 for a in $k = 12a - 14$

$k = 12 * 1 - 14$

$k = 12 - 14$

$k = -2$

3 0

2 years ago

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

So

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

So

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

So

$y_2'' + y_2' = -4cos2x-2sin2x$

So

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

So

$y_3'' + y_3' = -4sin2x - 2cos2x$

So

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

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2 years ago

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AlladinOne [14]

What do you mean by this?

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2 years ago

Part A Predict what will happen to the coordinates of a point (x, y) as the point reflects across a line to produce point (x', y

mylen [45]

The list attached shows the coordinate of (x, y) after transformation when reflecting over x, y-axis, and lines y = x, y = -x

<h3>What is geometric transformation?</h3>

It is defined as the change in coordinates and the shape of the geometrical body. It is also referred to as a two-dimensional transformation. In the geometric transformation, changes in the geometry can be possible by rotation, translation, reflection, and glide translation.

We have two lines of reflection:

y = x and

y = -x

The point is (x, y)

After reflecting over line y = x the coordinate (x, y) becomes:

(y, x)

After reflecting over the line y = -x the coordinate (x, y) becomes:

(-y, -x)

Thus, the list attached shows the coordinate of (x, y) after transformation when reflecting over x, y-axis, and lines y = x, y = -x

Learn more about the geometric transformation here:

brainly.com/question/16156895

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1 year ago