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3 years ago

How do you find an equivalent ratio?

1 answer:

5 0

It can be written two ways; for example, 1:10 or 1/10. Equivalent ratios are two ratios that express the same relationship between numbers. You can create equivalent ratios by multiplying or dividing both the numerator and denominator of a given ratio by the same number.

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The admission fee at an amusement park is $1.50 for children and $4 for adults. On a certain day, 250 people entered the park, a

pogonyaev

Answer:

100 children

150 adults

Step-by-step explanation:

c- children

a- adults

---------------

we get equations as per question:

1.5c+4a= 750
c+a= 250 ⇒ a= 250- c

1.5c +4(250-c)= 750 ⇒
1.5c- 4c+1000=750 ⇒
2.5 c= 250 ⇒
c= 100
a= 250-100 ⇒
a= 150

Children 100, adults 150

6 0

2 years ago

Mrs. Lane took a survey of the types of pants her students were wearing. She collected the data at the right. What percent of he

Nadya [2.5K]

Answer:

It is always important to go through the given problem first to get a concept of the requiremement. Then all the information's available from the question has to be noted down in such a manner that there would be no need to look at the question while solving.

Total number of students wearing jeans = 10

Total number of students wearing shorts = 9

Total number of students wearing capris = 2

Then the total number of students surveyed by Mrs Lane = (10 + 9 + 2)

= 21

Now percentage of students wearing shorts = (9/21) * 100

= (3/7) * 100

= 300/7

= 42.85 percent

So a total percentage of 42.85% of the students were wearing shorts.

Step-by-step explanation:

6 0

3 years ago

-3(6) + 2(3) - 18 = 6<br><br>Is this incorrect or correct and why ?

KiRa [710]

It is incorrect because when you times a negative by a positive. It is a negative number.

6 0

3 years ago

Solve by using cramers rule : 3x +2y = 1 , x-y = 2

Hoochie [10]

Answer:

its 3hdnndndjsjshnanajzh z

5 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago