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2 years ago

8

(2x^0y^2)^-3 multiply 2yx^3

1 answer:

hodyreva [135]2 years ago

3 0

Answer:

$\frac{x^{3} }{4y^{5} }$

Step-by-step explanation:

Simplifica la expresión

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A local hamburger shop sold a combined total of 500 hamburgers and cheeseburgers on Wednesday. There were 50 fewer cheeseburgers

Virty [35]

Hey there! I'm happy to help!

Let's call the hamburgers h and the cheeseburgers c.

h+c=500

c=h-50

Let's plug this value for c into the first equation to solve for h.

h+h-50=500

2h-50=500

Add 50 to both sides.

2h=550

Divide both sides by 2.

h=275

Therefore, 275 hamburgers were sold on Wednesday.

Have a wonderful day! :D

6 0

3 years ago

Which of the following best describe when you need to borrow during subtracting

Allisa [31]

Need the choices cant really answer your question

4 0

2 years ago

Need answer asap!!!!!<br><br> how would i solve this?

lesya [120]

The height of the building to the nearest tenth of a meter is; 13 m

<h3>How to make use of Cosine Rule?</h3>

The slant line from the top of Trevor's head to the base of the building is gotten from Pythagoras theorem;

15/x = sin 36°

x = 15/sin 36

x = 25.52 m

Angle between that slant line and base of building is;

90 - tan⁻¹(1.5/14) = θ

θ = 83.88°

Remaining angle of the bigger triangle is;

180 - (36 + 83.88) = 60.12°

Thus, if the height of the building is h, then;

h/sin 36 = 25.52/sin 60.12

h = 13 m

Read more about Cosine Rule at; brainly.com/question/4372174

#SPJ1

5 0

1 year ago

I need help with this will mark u as brainliest.

Amanda [17]

1. B
2. A
3. D
4. D
I'm not too sure about 1 but I'm sure about the others. Hope this helps!

7 0

3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago