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LuckyWell [14K]

3 years ago

12

The pet store started the day with 24 puppies and finished the day with 10 puppies. What was the percent change in the puppies a

t the store?
The answer choices are
1. 58%
2. 12%
3. 42%

2 answers:

Aleksandr [31]3 years ago

6 0

Answer:

58%

Step-by-step explanation:

11111nata11111 [884]3 years ago

3 0

A

14/24*100

Mark brainliest

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Why is rates, ratios ,and proportions important when working in the healthcare industry?

Romashka [77]

Learning the concepts of rates, ratios, and proportions is important to those in healthcare professions because this allows them to be accurate with the dosages that is to be given to their patients. Moreover, the changes in the amount of medicine that are to be made in accordance to the need of the patient may be easily determined.

5 0

3 years ago

A park in the shape of a right triangle is the set in the middle of 3 square rose gardens, with one garden along the side. Which

il63 [147K]

<u>Complete Question:</u>

A park in the shape of a right triangle is set in the middle of 3 square rose gardens, with one garden along each side. Which statement is true about the 3 gardens?

A) The area of Garden B minus the area of Garden C is equal to the area of Garden A.

B) The area of Garden B plus the area of Garden C is equal to twice the area of Garden A.

C) The area of Garden C minus the area of Garden B equals the area of Garden A.

D) The area of Garden C minus the area of Garden A is equal to twice the area of Garden B.

<u>Correct Answer:</u>

The area of Garden C minus the area of Garden B equals the area of Garden A.

Thus, (C) is the correct option.

<u>Step-by-step explanation:</u>

According to question , a right angle triangle with right angle at R and A is perpendicular , B is base and C is hypotenuse of the triangle.

One square garden along each side of triangle.

Area of square = Side × Side

So area of each square garden is :

$A ( A) = A^2\\B ( B) = B^2\\C ( C )= C^2$

According to Pythagoras theorem,

In a right angle triangle, square of hypotenuse is equal to sum of square of the remaining two sides.

By figure,

$A^2 + B^2 = C^2$

Area of garden A + area of garden B = Area of garden C

$A^2 = C^2 -B^2$

The area of Garden C minus the area of Garden B equals the area of Garden A.

Thus, (C) is the correct option.

8 0

3 years ago

Juan’s father is paying for a $45 meal. He has a 10%-off coupon for the meal. After the discount, an 8% sales tax is applied. Wh

leva [86]

Answer:

$43.74

Step-by-step explanation:

will pay= $45

10% of =$45×10/100=$4.5

so, $45-$4.5= $40.5

8% tax= $40.5×8/100 =$3.24

total= $43.74

7 0

3 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

3 years ago

in a small school, there are 60 boys and 80 girls. write down the ratio of the number of boys to the number of girls. give your

tester [92]

60:80
Simplest form = 3:4

3 0

3 years ago