Expand and simplify 9 (x-y)^2-9(x+y)(x-y)

The temperature at noon on a winter day was 8°C. At midnight, the temperature had dropped 15°C. What was the temperature at midn

klio [65]

All you have to look at is the word "dropped" and you know it's a subtraction problem. In this case, it would be 8 - 15.

8 - 15 = -7

It would be -7<span>°C at midnight. I hoped this helped! :)</span>

6 0

3 years ago

Dale keeps all of his pairs of shoes in his closet. select the number of shoes that deal could have in his closet. Mark all that

bekas [8.4K]

As it says he keeps 'pairs' of shoes so the answer has to be an even 'number' - ie B, C and E are all possible answers.

6 0

3 years ago

Read 2 more answers

In the figure to the right, AB = 7 1/2 m, BC = 2 1/4 m, and DE = 1 1/4 m. Find the perimeter of the figure if all angles are rig

Andreas93 [3]

Answer:

Perimeter=22 m

Step-by-step explanation:

Perimeter Of A Figure

Perimeter is the distance measured around a shape. If the figure is line-shaped, the perimeter can be obtained by adding the individual lengths of each segment around the shape.

The figure shown is surrounded by line segments. We only have to add them all to find the perimeter. But we don't need each individual length to do so. We may notice the following (given all angles are right):

The sum of HG+FE+DC equals AB. So the upper and lower lengths are twice AB, or equivalently: 2*7 1/2 m =15 m

It can also be noted that AH+GF=BC+DE=2 1/4+1 1/4 = 3 1/2 m. It means that the two lateral lengths are twice this value: 2* 3 1/2 = 7 m

Thus, the total perimeter is 15 m + 7 m = 22 m

4 0

3 years ago

Cole got 100 for his birthday. He can either save the money or buy a tennis racket with it. What is the opportunity cost if he b

hodyreva [135]

Cole can learn how to play tennis. But if he saved the money, he can save up and get something bigger, or save up for college. There is only so much he can do with a tennis racket except for using it to learn a new sport.

8 0

3 years ago

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