Someone help me find mCD please

X-5y=-15x−5y=−15x, minus, 5, y, equals, minus, 15 Complete the missing value in the solution to the equation. (-5,(−5,left paren

Y_Kistochka [10]

Answer:

<h2>The missing value is 2. The coordinate will be (-5, 2)</h2>

Step-by-step explanation:

The question is not properly written. Find the correct question below.

If x – 5y = -15 . Complete the missing value in the solution to the equation (-5, ____)

Let the coordinate of the variables be (x, y). Comparing the coordinates (x, y) with the given coordinate (-5, __), we will discover that x = -5. To get the y coordinate, we will substitute x = -5 into the given expression as shown;

If x – 5y = -15

-5 - 5y = -15

Adding 5 both sides

-5-5y+5 = -15+5

-5y = -10

Dividing both sides by -5;

-5y/-5 = -10/-5

y = 2

<u>Hence the missing value in the solution of the equation is 2</u>. The coordinate will be (-5, 2)

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

In art class,Kerion is coloring a pattern where half of the squares on her paper are being colored blue. Of the blue squares,1/3

kolbaska11 [484]

100% of Kerion's paper squares are:
(100% / 100%) = 1
Half of the squares of your paper are colored blue:
(1) / (2) = 1/2
Of the blue squares, 1/3 of them will also have stripes:
(1/2) * (1/3) = 1/6
answer
A fraction of (1/6) squares will be blue with strips

7 0

3 years ago

Plz help me with this

jeyben [28]

Only the last two options are true.

The first one is false, because 4.5 is the value of the maximum of f(x), not the point where it is reached.

The second one is false, because g(x) has a maximum of 9, so it is a downward-facing parabola (just like f(x)), so it doesn't have a minimum.

The third one is true, because the maximum value of f(x) is 4.5, and the maximum value of g(x) is 9, which is twice the maximum of f(x)

The last one is false (see point 2).

3 0

3 years ago

Read 2 more answers

The function f(3) = 50.000(0.975)", where x represents the underwater

valentina_108 [34]

9514 1404 393

Answer:

316 lumens/m²

Step-by-step explanation:

We assume your intensity function is ...

f(x) = 50,000(0.975^x) . . . . x meters below the surface

__

Use 200 for x and do the arithmetic.

f(200) = 50,000(0.975^200) ≈ 316 . . . . lumens/m²

6 0

3 years ago