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4 years ago

Find two numbers, if their sum is (−1/3) and their difference is 18. PLEASE HELP

1 answer:

4 0

X+y=-1/3
x-y=18
solve the system of equation:
equation 1 + equation 2: 2x=17&2/3=16&5/3
x=8&5/6
y=-1/3 -8&5/6=-8&7/6=-9&1/6

x= $x=8\frac{5}{6} , y=-9 \frac{1}{6}$

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Please help with the 2 questions ive attached, i dont understand it at all

vesna_86 [32]

Answer:

see explanation

Step-by-step explanation:

(a)

Consider the denominator

To obtain 5xy from x , multiply by 5y

Thus numerator is multiplied by 5y, that is

$\frac{1}{x}$ = $\frac{5y}{5xy}$

(b)

Consider the numerator

To obtain 5ab² from ab, multiply by 5b

Thus denominator is multiplied by 5b, that is

$\frac{ab}{2c}$ = $\frac{5ab^2}{10bc}$

3 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

The table represents a linear function. Find the values of A and B.

bazaltina [42]

Answer:

The answers are that a = -5 and b = 1

Step-by-step explanation:

In order to find A and B, we first need to find the equation of the line. We can do this by using two ordered pairs and the slope formula. For the purpose of this activity, I'l use (0, 5) and (-3, 11)

m(slope) = (y2 - y1)/(x2 - x1)

m = (11 - 5)/(-3 - 0)

m = 6/-3

m = -2

Now that we have this we can model this using point-slope form.

y - y1 = m(x - x1)

y - 5 = -2(x - 0)

y - 5 = -2x

y = -2x + 5

Now that we have the modeled equation we can use the ordered pair (a, 15) to solve for a.

y = -2x + 5

15 = -2(a) + 5

10 = -2a

-5 = a

And we can also solve for b using the ordered pair (2, b)

y = -2x + 5

b = -2(2) + 5

b = -4 + 5

b = 1

4 0

3 years ago

Please help!!

sergij07 [2.7K]

Answer: B
Explanation: It’s linear because as y increases x also increases however it’s weak because there are a few anomalies/ outliers

6 0

2 years ago

2(9+10r)-14 please help

Fittoniya [83]

Answer:

$20r+4$

7 0

3 years ago