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3 years ago

[Simplify Radical Equations].

2 answers:

4 0

It'd help both of us if you'd please share whatever work you've done, right or wrong. Similar or same basic principles apply to all of these problems. You could learn the pattern and apply it yourself.

1) √125 could be broken down into two factors: sqrt(5*25) = 5sqrt(5) (ans.)

5)√216 Think: perfect square factors include 1, 4, 9, 16, 25, 36, 49, 81, 100, 121, etc. Which of these perfect squares divide into 216 with no remainder? We want the largest such square. It is 36. Therefore,
sqrt(216) = sqrt(36*6) = 6sqrt(6) (answer)

9)√147 Think: what is the largest perfect square factor of 147? Look at the list above (Problem 5). How about 49? Then sqrt(147) = sqrt(49)*sqrt(3), or 7sqrt(3).

Please choose 2 of the remaining problems. Follow my examples. Share your work. I'll return if at all possible and continue to help you.

Ludmilka [50]3 years ago

3 0

sqrt(216) = sqrt(36*6) = 6sqrt(6)

thats for number five

You might be interested in

3) The temperature was -27°F and then it dropped another 7 degrees. What was the temperature after the drop?

kirill [66]

Answer:

-34°F

Step-by-step explanation:

-27 - 7 = -34

27 + 7 = 34 = -34

I hope this helps! have a wonderful rest of ur day :)

8 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Help with this please

Paha777 [63]

I think it is B
hope this helps

6 0

3 years ago

1) (1 Point) : If-18 - x=9, then -5 +6x = ?

TiliK225 [7]

-167, because if you solve -18-x=9, x will be equal to -27. When you put -27 in place of the x in -5+6x, you get -167. Have a nice day!

3 0

3 years ago

Pls help me !!!!!!!!!!!!!!!!!!!!!!!!!!

marysya [2.9K]

Answer:

1. y^12/y^12=1

2. q^15*1/q^15=1

3. (7(123456.789)^4)^0=1

4. 2^2*1/2^5=1/2^3

5 (x^41/y^15)*(y^15/x^41)=1

Step-by-step explanation:

1. when dividing exponents, subtract the exponents. y^12/y^12=y^(12-12)=y^0=1

2. you do the same as #1

3. Any number to the power of 0 is always 1

4. 1/2^3=1/8

5. since the numerator and denominator is the same, the divide to equal 1

Hope I helped

8 0

3 years ago