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3 years ago

Explain why each pair does not belong 2, 28 4,14 6,9 and 7,8?

1 answer:

4 0

2 and 28 do belong together 4 and 14 do not belong together because there is nothing to multiply 4 by to get 14.6 and 9 do not belong together because as I said last time there is nothing to multiply etc........

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Which expressions do you get when you eliminate the negative exponents of .........

kvasek [131]

Answer:

D. 4(a^2)(a^3)/16(b)(b^2)

Step-by-step explanation:

5 0

3 years ago

Choose the correct answer: 15 is 3% of<br> (A) 1.5<br> (B) 2.5<br> (C) 500 <br> (D) 5000

eduard

Answer:

c) 500

Step-by-step explanation:

3% = 0.03

multiply the decimal from of 3% by 500:

500 x 0.03

solve:

500 x 0.03 = 15

3 0

3 years ago

If 6 cups of cheese serves 30 people how cups of cheese serve 50

bezimeni [28]

6 cups of cheese servings times 5 = 30
10 cups of cheese servings time 5 = 50
So , it's 10 cups of cheese servings.

4 0

3 years ago

Read 2 more answers

Emily's family loves to work together in the garden. They have a slight preference for flowers, as 60% percent of their plants a

Phantasy [73]

20 because 100 plants would have 60 flowers and 40 vegetables so 50% which is half would be 30 flowers and 20 vegetables

4 0

4 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

3 years ago