<img src="https://tex.z-dn.net/?f=%28%20-%209%29%20%7B%20%7D%5E%7B2%7D%20%20-%20%20%5Csqrt%5B3%5D%7B8%3F%7D%20" id="TexFormula1"

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pshichka [43]

3 years ago

title="( - 9) { }^{2} - \sqrt[3]{8?} " alt="( - 9) { }^{2} - \sqrt[3]{8?} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

dlinn [17]3 years ago

6 0

Answer:

Step-by-step explanation:

(-9)² - ∛8

Square the -9. (-x)² = x².

81 - ∛8

Calculate the cube root of 8.

81 - 2

Subtract.

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Leah is 2 less than 3 times Rachel's age. 3 years from now, Leah will be 7 more than twice Rachel's age. How old will Rachel be

Irina-Kira [14]

Answer:

Step-by-step explanation:

Let's call Leah's age l and Rachel's age r. We can write:

l = 3r - 2 (1)

l + 3 = 2(r + 3) + 7 (2)

Substituting (1) into (2) we get:

(3r - 2) + 3 = 2(r + 3) + 7

3r + 1 = 2r + 13

r = 12

In 3 years Rachel will be 12 + 3 = 15 years old.

8 0

3 years ago

Karan needs to choose between two gym plans. He can either pay a $ 1 5 0 $150dollar sign, 150 joining fee and a $ 1 0 $10dollar

stira [4]

Calculation for plan A:

Let numbe of months = x

Given that $10 is the monthly fee.

Given that $150 is the joining fee.

Then cost equation for plan A is given by:

A=10x+150

Calculation for plan B:

Let numbe of months = x

Given that $30 is the monthly fee.

Given that $50 is the joining fee.

Then cost equation for plan B is given by:

B=30x+50

Now we have to find on what month will the cumulative costs of the plans be equal.

So we will just set both equations equal to each other

10x+150=30x+50

150-50=30x-10x

100=20x

5=x

Hence it will take 5 months to make cost of both plans equal.

To find the total cost plug x=5 into any of above equations:

A=10x+150=10*5+150=50+150=200

Hence final cost will be $200.

8 0

3 years ago

Megan bikes west to get from her apartment to school after school she bikes 8 miles north to her friends house how far is megans

snow_tiger [21]

The complete question: Megan bikes 6 miles west to get from her apartment to school. After school, she bikes 8 miles north to her friend's house. How far is Megan's apartment from her friend's house, measured in a straight line?

So,Megan's apartment, the school, and Megan's friend's house form a right triangle with legs 6 miles and 8 miles as you can see in the attached picture; the hypotenuse of the right triangle is the straight distance is from Megan's apartment to her friend's house. To find that distance we are going to use the Pythagorean theorem:
 $h= \sqrt{(6mi)^2+(8mi)^2}$
$h= \sqrt{36mi^2+64mi^2}$
$h= \sqrt{100mi^2}$
$h=10mi$

We can conclude that distance between Megan's apartment and her friend's house is 10 miles.

4 0

3 years ago

What is the equation of the graph below in slope intercept form? You MUST reduce your slope to its simplest answer

Sedaia [141]

Answer:

53.6

Step-by-step explanation:

7 0

3 years ago

NO LINKS!!! Please help me with parts b and c.

bekas [8.4K]

9514 1404 393

Answer:

b. A × B = GCF(A, B) × LCM(A, B)

c. 15 = GCF × LCM / 20

Step-by-step explanation:

b. The products of the numbers in the table are 80, 108, 216, 45.

The products of the corresponding GCF and LCM are 80, 108, 216, 45.

Conjecture: the product of the numbers is equal to the product of the GCF and LCM.

c. GCF×LCM = 5×60 = 300 = A×B = 20×B

B = 300/20 = 15

The other number is 15, found by using the conjecture of part b.

_____

<h3>Additional comment</h3>

If you think for a minute about what the GCF and LCM are, you realize the relationship discussed in this problem must be the case. Consider the factors of two numbers A and B:

A = (factors in common with B) × (factors unique to A)

B = (factors in common with A) × (factors unique to B)

The GCF, by definition, are (factors in common between A and B).

The LCM, by definition, is the product of unique factors:

(common factors) × (factors unique to A) × (factors unique to B)

Then the product of the LCM and the GCF is ...

GCF × LCM =

(factors in common) × ((factors in common) × (unique to A) × (unique to B))

Using the associative and commutative properties of multiplication, we can rearrange this product to be ...

((factors in common)×(unique to A)) × ((factors in common)×(unique to B))

= A × B

Personally, I use a kind of diagram to represent the factorization of A and B and their LCM and GCF:

[unique to A (common] unique to B)

Then A = factors in [ ], and B = factors in ( ). The stuff in (common] is the GCF, and the overall product is the LCM.

Using the example of part c, this would look like [4 (5] 3), so A = 4·5 = 20 and B = 5·3 = 15. The GCF is 5, and the LCM is 4·5·3 = 60.

The end parts [unique to A( and ]unique to B) can have no common factors. Any common factors must reside in the (common] part.

6 0

3 years ago

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