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3 years ago

7

Please help me is for my daughter

1 answer:

stich3 [128]3 years ago

7 0

Answer:

A

Step-by-step explanation:

You might be interested in

The perimeter of a rectangle Is 108 centimeters. If the length of the rectangle is five times the width, what are the dimensions

vampirchik [111]

Answer: L=45, W=9

Step-by-step explanation:

Perimeter P=108 so if L=5W,

P=2(L+W)=2(5W+W)=2(6W)=12W

108=12W

W=108/12=9

L=5*9=45

4 0

3 years ago

a set of five numbers has a mode of 24 a median of 21 a mean of 20 . work out what the numbers could be

SSSSS [86.1K]

Answer:

Step-by-step explanation:

The mode is 24 so we know that at least 2 of the 5 numbers are 24.

The mean is 20 so we know that the 5 numbers add up to 100.

The median is 21 so the third number is 21.

100-(24+24+21)= 31

Now, I am pretty certain that the last 2 numbers can be anything as long as they add up to 31 AND they are not also 21 because 24 then would no longer be the mode.

E.g. 24, 24, 21, 11 & 20

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2

kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed $a\leq x \leq b$ and in your case $\nexists x\in\mathbb{Z}: a\leq x\leq b$ for $a>b$

In fact, considering a set $S$ we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing$ that satisfy $S = S \cup \varnothing$

This means that, by definition $\sum_{x=a}^b \varnothing = 0$

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

we have other problems. Actually, this case is really bad.

Note that $\cos^2(\infty)$ has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$ , the cosine function oscillates between $[-1, 1]$ , and therefore it is undefined. Thus, we cannot evaluate

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0

3 years ago

Solve for x. 3x+2 = 8x-1

Flauer [41]

So, your problem is ((3x+2))/4) = ((8x-1)/5)?

If so, X= 14/17

6 0

4 years ago

(2x^3+9x^2+x-12)/(x+4) simplify

neonofarm [45]

Answer:

<h2> $2 {x}^{2} + x - 3$ </h2>

Step-by-step explanation:

$\frac{2 {x}^{3} + 9 {x}^{2} + x - 12}{x + 4}$

Write $9 {x}^{2}$ as a sum

$\frac{2 {x}^{3} - 2 {x}^{2} + 11 {x}^{2} + x - 12}{x + 4}$

Write X as a sum

$\frac{2 {x}^{3} - 2 {x}^{2} + 11 {x}^{2} - 11x + 12x - 12 }{x + 4}$

Factor out $2 {x}^{2}$ from the expression

$\frac{2 {x}^{2}(x - 1) + 11 {x}^{2} - 11x + 12x - 12 }{x + 4}$

Factor out 11x from the expression

$\frac{2 {x}^{2} (x - 1) + 11x(x - 1) + 12x - 12}{x + 4}$

Factor out 12 from the expression

$\frac{2 {x}^{2}(x - 1) + 11(x - 1) + 12(x - 1) }{x + 4}$

Factor out X - 1 from the expression

$\frac{(x - 1)(2 {x}^{2} + 11x + 12)}{x + 4}$

Write 11x as a sum

$\frac{(x - 1)(2 {x}^{2} + 8x + 3x + 12)}{x + 4}$

Factor out 2x from the expression

$\frac{(x - 1)(2x(x + 4) + 3x + 12)}{x + 4}$

Factor out 3 from the expression

$\frac{(x - 1)(2x(x + 4) + 3(x + 4)}{x + 4}$

Factor out X + 4 from the expression

$\frac{(x - 1)(x + 4)(2x + 3)}{x + 4}$

Reduce the fraction with X + 4

$(x - 1)(2x + 3)$

Multiply the parantheses

$2 {x}^{2} + 3x - 2x - 3$

Collect like terms

$2 {x}^{2} + x - 3$

Hope this helps...

Good luck on your assignment...

4 0

3 years ago