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2 years ago

Can you please help me with number 3

Mathematics

1 answer:

love history [14]2 years ago

5 0

Answer:

17in

Step-by-step explanation:

a^2+b^2=c^2 <---- formula

so...

c^2=8^2+15^2

c^2= 289 (but you only want c by its self so you need to square root it on both sides because what you do on one side you do the same on the other)

$\sqrt{c^{2} }$ = $\sqrt{289}$

c= 17 inch

thus the answer is 17 inches

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2x^4–5x^3+x^2+3x+2=? X=5

DanielleElmas [232]

Answer:

$2x^4–5x^3+x^2+3x+2=$

$2 \times 5^4–5 \times 5^3+5^2+3 \times 5+2$

= 2 × 625 – 5 × 125 + 25 + 3 × 5 + 2

=1250 – 625 + 25 + 15 +2

= 1292 – 625

= 667

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3 years ago

HELP ASAP Use the data below

beks73 [17]

Answer:

Part A:

-Minimum: 10

-Q1: 17.5

-Median: 30

-Q3: 42.5

-Maximum: 50

Step-by-step explanation:

Part B: IQR= 25

This shows that the data varies for 25 different numbers. That HALF of the data is between 25 numbers.

Part C: Using a box-and-whisker plot you can interpret the different values. Minimum is the first dot (10), connected to the first line (Q1 which is 17.5), connected by a box to the median (30), connected by a box to the third line (Q3 which is 42.5), connected to the last dot which is the maximum (50). And IQR is Q3-Q1, so 42.5-17.5 which is 25.

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3 years ago

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faltersainse [42]

Answer:

Step-by-step explanation:

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2 years ago

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il63 [147K]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

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

3 years ago