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2 years ago

Please answer this question ‍♀️

Mathematics

1 answer:

krek1111 [17]2 years ago

5 0

Answer:

sum is 1,2,4,8,16,32,64,128,256,512

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Help me guys pls pls pls i need a 70 or higher

Elis [28]

Answer:

x - 36 ≥ 24

Ms. Estrada will bake at least 60 cookies

Step-by-step explanation:

8 0

3 years ago

A candy store grab bag contains 10 pieces of sour candy, 12 pieces of chocolate candy, and 6 pieces of sweet candy. If you selec

harkovskaia [24]

Step-by-step explanation:

so we're making two draws *with* replacement (this is important)

step 1: for the first draw, it wants the probability of getting a sour candy. to calculate this:

(# of sour candy) / (total # of candy)

step 2: for the second draw, it wants the probability of *not* getting a sour candy. to calculate this, you can calculate 1 - (the probability form part 1).

step 3: to find the probability of both events happening together, simply multiply the probabilities from part 1 and 2 together

side note: for step 2, you can only do this because the candy is being replaced. if there were no replacement, you'd have to re-calculate (# of non-sour candies) / (total after the first candy is drawn)

4 0

2 years ago

Read 2 more answers

Find the area of the triangle shown 17ft 15ft 8ft

bonufazy [111]

Answer:

127 ft

BASE * HEIGHT

8 0

2 years ago

What value is true of x makes this equation true 0.7x-5=0.2x+1

Sever21 [200]

Answer:

x=12

Step-by-step explanation:

First you make sure x is on both sides of the equation. So you do 0.7x-5-0.2x=0.2x+1-0.2x. Which just simplifies to 0.5x-5=1. You make sure x is the only thing on that side of the equation so you do 0.5x-5+5=1+5 which simplifies to 0.5x=6. Multiply the equation times 2 to just have x. x=12. The value that makes true of x is 12.

7 0

3 years ago

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

Now for partial fractions: you're looking for constants a, b, and c such that

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$

7 0

2 years ago