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2 years ago

PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER

Mathematics

2 answers:

jekas [21]2 years ago

7 0

The correct answer is D

V125BC [204]2 years ago

7 0

Let's do it in steps:
7³ = 7 x 7 x 7
7 x 7 = 49
49 x 7 = 343

The correct answer is D) 343

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What is the solution set of x2 + y2 = 26 and x − y = 6? A. {(5, -1), (-5, 1)} B. {(1, 5), (5, 1)} C. {(-1, 5), (1, -5)} D. {(5,

Rus_ich [418]

He two equations given are
x^2 + y^2 = 26
And
x - y = 6
x = y +6
Putting the value of x from the second equation to the first equation, we get
x^2 + y^2 = 26
(y + 6) ^2 + y^2 = 26
y^2 + 12y + 36 + y^2 = 26
2y^2 + 12y + 36 - 26 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
y^2 + y + 5y + 5 = 0
y(y + 1) + 5 ( y + 1) = 0
(y + 1)(y + 5) = 0
Then
y + 1 = 0
y = -1
so x - y = 6
x + 1 = 6
x = 5
Or
y + 5 = 0
y = - 5
Again x = 1
So the solutions would be (-1, 5), (1 , -5). The correct option is option "C".

4 0

2 years ago

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A hot tub pump circulates 3.9 cubic feet of water per minute.how many cubic inches of water does it circulate each second

Basile [38]

Answer:

0.78 cubic inches/sec

Step-by-step explanation:

you take 3.9 cubic feet and multiply it by 12 to convert it into inches and you get 46.8 cubic inches. then you divide by 60 to convert minutes to second and you get 0.78. hope that helps

5 0

2 years ago

Read 2 more answers

How many 5 digit numbers are even if repetition is not allowed?

shtirl [24]

23308 is the answer of how many is even with no repitition

7 0

2 years ago

Challenge You have 2 different savings accounts. For Account A, the simple interest earned after months is $. For Account B,

Debora [2.8K]

Answer and Step-by-step explanation:

The computation is shown below;

As we know that

The simple interest is

Simple interest = Principal × rate × time

For Account A, the principal would be

2 = Principal × 3.2% × 3

So, the principal is $20.83

For account B, the principal would be

38.50 = Principal × 2.2% × 30

So, the principal is $58.33

Now the interest for both the accounts are as follows

For account 1

= $20.83 × 3.2%

= $0.67

And, for the account 2

= $58.33 × 2.2%

= $1.28

As it can be seen that the account 2 has the highest interest

The same is to be considered

7 0

2 years ago

Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago