All categories

3 years ago

Help me please I just need math and that's all for math

Mathematics

2 answers:

belka [17]3 years ago

7 0

3 1/4 9 1/2

hope it helpedd

Sliva [168]3 years ago

5 0

1. 3 1/4
2.9 1/2 hope it helped

You might be interested in

Equivalent fractions for 9/11, 15/17, and 2/3 using the least common denominator.

TiliK225 [7]

The (small) challenge here is to determine what the LCD is. Since 11, 17 and 3 are all prime, we find the LCD by multiplying them together: 561.

Then 9/11= 459/561; 15/17 = 495/561; and 2/3 = 375/561.

8 0

3 years ago

The graph of the equation y = į x+2 is displayed.

Oksi-84 [34.3K]

Answer:

first

third and

fourth

7 0

3 years ago

Nicole is 56 inches tall and has an 84 inch shadow. Julie is standing next to Nicole and has a 96 inch shadow. How tall is Julie

ki77a [65]

I believe it’s 68 Inches

5 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

A word processing program requires a user to enter a 7-digit registration code made up of the digits 1,2,4,5,6,7 and 9. Each num

ahrayia [7]

For the first digit there are 7 choices. For the second digit there are 6 choices (because we can't use the same one again). And so on, until there is only one choice for the last digit.

The number of possible codes is this:
7*6*5*4*3*2*1 = 7! = 5040

6 0

3 years ago