Which point on the number line represents two and one-half degrees below zero?

The sum of 3 consecutive integers is 47 more than the first integer. Find the three integers.

ss7ja [257]

Answer:

22,23,24

Step-by-step explanation:

Well, you need to first find two consecutive integers that add up to 47, because that is the difference between the first digit and the total. This happens to be 23 and 24. Once you get these two numbers, the 22 comes intuitively, as you have to add 23 and 24 to get a difference between 22 and (22+23+24).

7 0

2 years ago

Which expressions are equivalent to -90 – 60w?

Feliz [49]

Answer:

-30 (2w + 3) is the only one I can figure out.

I hope this helped tho!! :]

3 0

2 years ago

I need answers fast plsss

Nonamiya [84]

Answer:

1/6

Step-by-step explanation:

the form you're using is slope-intercept. That is

y=mx+b

where y= y coordinate

m=slope

x= x coordinate

and b= y intercept

so, m in that equation is 1/6, so the slope is 1/6

Hope this helps!

6 0

2 years ago

A grade has 81 girls and 72 boys. The grade is spilt into groups that have the same ratio of boys to girls as the whole grade. H

MariettaO [177]

There are 81 girls to every 72 boys in total, which can be represented by 81/72, which put into decimal form, is 1.125. If you then take the number of boys in the group, 16, and multiply it by this number (because the problem states that the ratio is constant) you can find the number of girls in the group.

16*1.125=18

So there are 18 girls in a group with 16 boys.

4 0

3 years ago

Read 2 more answers

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

3 years ago