Find the circumference of a circle if the area is 300 square centimeters

Answers

Answer 1

Answer: 61 is the answer i think

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On Monday, your bank account balance was −$12.92. Because you didn't realize this, you wrote a check for $30.41 for groceries. What is the new balance in your account?

Answers

-43.33 because you do -12.92 minus 30.41 and then you get -44.33

Is log2+log18 equal to log30+log6

Answers

$No,\ is's\ not\ because:\n\nlog2+log18=log(2\cdot18)=log36\n\nand\n\nlog30+log6=log(30\cdot6)=log180\n\nlog36\neq log180$

No it's not log2+log18= 1.55, while log30+log6=2.25

Which of the following shows the correct factorization of x3 - 5x2 - 14x?A. x(x + 7)(x + 2)

B. x(x - 7)(x - 2)

C. x(x + 7)(x - 2)

D. x(x - 7)(x + 2)

Answers

So, just a note of formatting, to indicate something is an exponent of something else, use the carat (^) sign
for instance
x^3 means x cubed. x3 means 3x.
Anyways, though, let's factor.
You can factor out an x from each variable.
x(x^2-5x-14)
Now, to factor this, think of what will add to -5, and multiply to -14.
-7 and 2 work.
So, this factors out to
x(x-7)(x+2)
Choice D is the correct factorization.

Tell whether 30 is divisible by 2,3,5,6 or 9

Answers

The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30 .

9 is not a factor of 30 .

30/3=10
30/2=15
30/5=6
30/6=5
30/9=3.333333333333

Solve for x and y in the following set of equation 6x+5xy-5y=8
X+y=3

Please help this is important

Answers

STEP 1:

$6x+5xy-5y=8\n \n 6x+5y\left( x-1 \right) =8\n \n 5y\left( x-1 \right) =8-6x\n \n 5y\left( x-1 \right) =2\left( 4-3x \right) \n \n \frac { 1 }{ 5\left( x-1 \right) } \cdot 5y\left( x-1 \right) =2\left( 4-3x \right) \cdot \frac { 1 }{ 5\left( x-1 \right) } \n \n y=\frac { 2\left( 4-3x \right) }{ 5\left( x-1 \right) }$

STEP 2:

$x+y=3\n \n y=3-x$

STEP 3:

This means that...

$3-x=\frac { 2\left( 4-3x \right) }{ 5\left( x-1 \right) } \n \n \left\{ 5\left( x-1 \right) \right\} \left( 3-x \right) =2\left( 4-3x \right) \n \n \left( 5x-5 \right) \left( 3-x \right) =8-6x\n \n 15x-5{ x }^( 2 )-15+5x=8-6x\n \n 20x-5{ x }^( 2 )-15=8-6x\n \n 5{ x }^( 2 )+8-6x+15-20x=0\n \n 5{ x }^( 2 )-26x+23=0$

STEP 4:

Which means that...

$\n \frac { 1 }{ 5 } \cdot \left( 5{ x }^( 2 )-26x+23 \right) =0\cdot \frac { 1 }{ 5 } \n \n { x }^( 2 )-\frac { 26 }{ 5 } x+\frac { 23 }{ 5 } =0\n \n { x }^( 2 )-\frac { 26 }{ 5 } x=-\frac { 23 }{ 5 }$

Which means that...

$\ \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )-{ \left( \frac { 13 }{ 5 } \right) }^( 2 )=-\frac { 23 }{ 5 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )-\frac { 169 }{ 25 } =-\frac { 23 }{ 5 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )=-\frac { 115 }{ 25 } +\frac { 169 }{ 25 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )=\frac { 54 }{ 25 }$

Which means that...

$\n \n x-\frac { 13 }{ 5 } =\pm \frac { \sqrt { 54 } }{ 5 } =\pm \frac { 3\sqrt { 6 } }{ 5 } \n \n \therefore \quad x=\frac { 13 }{ 5 } \pm \frac { 3\sqrt { 6 } }{ 5 }$

STEP 5:

When:

$x=\frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 }$

.........

$y=3-\left( \frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 } \right) \n \n y=\frac { 15 }{ 5 } -\frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 } \n \n y=\frac { 15-13-3\sqrt { 6 } }{ 5 } \n \n y=\frac { 2-3\sqrt { 6 } }{ 5 }$

STEP 6:

When :

$x=\frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 }$

...........

$y=3-\left( \frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 } \right) \n \n y=\frac { 15 }{ 5 } -\frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 } \n \n y=\frac { 15-13+3\sqrt { 6 } }{ 5 } \n \n y=\frac { 2+3\sqrt { 6 } }{ 5 }$

What is the answer to this question

Answers

Answer:

Step-by-step explanation:

It is 3 because you multiply the powers when in brackets

The answerrrrr is 3 (:

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