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3 years ago

Solve the equation using square roots . X^2 +20=4

Mathematics

2 answers:

Savatey [412]3 years ago

7 0

X²+20=4
X²=-16
No number squared can be a negative, so X=4i

vladimir1956 [14]3 years ago

7 0

Answer:<em><u> no real number solutions</u></em>

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PLEASE HELP ME!! THANK YOU!!

Alik [6]

Answer:

y $\leq$ 4,

y ∈ [-∞, 4]

Step-by-step explanation:

Remember, the range of the function is all possible output values or y values.

As per the graph, it appears that the output, the y value, is less than or equal to 4. Hence:

y $\leq$ 4

y ∈ [-∞, 4]

is the range, both are the same, they are just different ways of writing it.

6 0

3 years ago

What is the missing number in this sequence: –1, –2, –4, , –16, –32

kipiarov [429]

Answer:-8

Step-by-step explanation: you add -1+ -1, to get -2, then -2+-2, to get -4 and so forth.

5 0

3 years ago

Tina is shopping for school clothes. She realizes that good sales can help her to get more clothes and keep within her budget. S

podryga [215]

Answer:

$23.952, since money is always rounded up $23.96

Step-by-step explanation:

the first step is to get the total value of her purchases, thus we add 16.99 + 12.95 to get 29.94

since she will receive a 20% discount that means that she will be paying 80% of the total value

next we set up the ratio as follows:

portion paid / total = x/29.94 = 80/100

solving the ratio gives you 23.952

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

For a large data set of a average student grades, the third quartile Q3 is found 78.5 . What does the mean when we say that 78.5

xz_007 [3.2K]

Step-by-step explanation:

We are told that for a large data set of a average student grades, the third quartile Q3 is found 78.5.

Since we know that third quartile represents the number such that 75% of the data is less than this number. Third quartile is the middle value between median and the highest value of a data set. Third quartile is also known as upper quartile. Third quartile splits lower 75% data from highest 25% of data.

Therefore, third quartile represents that 75% of student grades are less than 78.5 and 25% of students grades are greater than 78.5.

3 0

4 years ago