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2 years ago

I WILL GIVE BRAINLIEST!!!

Mathematics

1 answer:

adell [148]2 years ago

6 0

Answer:

2 by 4 by 678

Step-by-step explanation:

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A line through slope -2 and y-intercept 4

m_a_m_a [10]

Answer:

y = -2x + 4

Step-by-step explanation:

When given a slope, and y intercept, this means the question is asking us to write the equation in slope-intercept form, which is y = mx + b.

Plug in the values.

y = -2x + 4

7 0

3 years ago

Selena bought a TV for her bedroom. The original price was $299.99, but she bought it on sale for 30% off. Selena also paid 15%

ycow [4]

I believe $241.50 would be your answer

8 0

2 years ago

In the figure above if PQRS is a square, what is the value of a?

yanalaym [24]

Answer:

sino me quivoco es 1/53 espero aber ayudado

6 0

2 years ago

Jenny multiplies the square root of her favorite positive integer by $\sqrt{2}$. Her product is an integer. a) Name three number

Dmitry [639]

Answer:

Part a) $2,50,18$

Part b) When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

Step-by-step explanation:

Let

x-------> the favorite positive integer

Part a)

1) For $x=2$

$\sqrt{2}*\sqrt{2}=\sqrt{4}=2$ -----> the product is an integer

The number $x=2$ could be Jenny favorite positive integer

2) For $x=50$

$\sqrt{50}*\sqrt{2}=\sqrt{100}=10$ -----> the product is an integer

The number $x=50$ could be Jenny favorite positive integer

3) For $x=18$

$\sqrt{18}*\sqrt{2}=\sqrt{36}=6$ -----> the product is an integer

The number $x=18$ could be Jenny favorite positive integer

Part B)

1) For $x=2$

$\sqrt{2}/\sqrt{2}=\sqrt{1}=1$ -----> the result is an integer

2) For $x=50$

$\sqrt{50}/\sqrt{2}=\sqrt{25}=5$ -----> the result is an integer

3) For $x=18$

$\sqrt{18}/\sqrt{2}=\sqrt{9}=3$ -----> the result is an integer

Therefore

When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

4 0

2 years ago

Write in simplest form:<br> <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24a%5E%7B10%7Db%5E%7B6%7D%7D" id="TexFormula1" ti

Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

$10=3+3+3+1$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

:Done

6 0

2 years ago