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2 years ago

1 point What is 20% of 60? Type your answer...

Mathematics

2 answers:

Lilit [14]2 years ago

8 0

Answer: 12

Step-by-step explanation:

alekssr [168]2 years ago

7 0

Answer:12

60×.20= 12

I hope this is good enough:

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Handsome Jack is buying a pony made of diamonds. The price of the pony is P dollars, and Jack also has to pay

Leya [2.2K]

Answer:

.25(P), or P/4

Step-by-step explanation:

Depending on what P, the answer should be .25(P)

7 0

3 years ago

In a U.S poll 8 out of 12 citizens said they were happy with the job Obama is doing. If 126 people were surveyed. How many peopl

Illusion [34]

You could solve this using a proportion or just by common sense.

Proportion: 8/12 = x/126

Cross multiply: 12x = 1008

1008/12 = 84

Or the common sense route is just dividing 12 by 8 then multiplying the answer by 126. Either way, you get the same answer. Hope this helps! :)

~Ash

6 0

3 years ago

Read 2 more answers

Suppose that you have he option to buy an item with either cash or credit. Under which condition would you use credit?

Shkiper50 [21]

If they didn't take cash

8 0

2 years ago

Read 2 more answers

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago

1) Given h(x)=x-5 <br> a) Determine h^2<br> b)determine h^3

ElenaW [278]

Answer:

A) x^2-10x+25 B)X^3-15x^2+75x-125

Step-by-step explanation:

H^2=(x-5)(x-5)=x^2-10x+25

H^3=(x-5)(x-5)(x-5)=X^3-15x^2+75x-125

6 0

3 years ago

1 point What is 20% of 60? Type your answer...​

1 point What is 20% of 60? Type your answer...