could anyone who me how to solve these problems 1. x^3=49x 2. 5y^2=80 3. 3x^2+6x+3=0 4. r^2-34-70=0?

Answer 1

1. x^3 = 49x
x^3/x = 49x/x Divide both sides by x
x^2 = 49 the result. Now take the square root of both sides
x = 7

2. 5y^2 = 80
1/5(5y^2) = (1/5) * 80 Divide both sides by 5
y^2 = 16 The result. Now take the sqrt of both sides
y = 4

3. 3x^2 +6x + 3 = 0
Use the quadratic formula to solve this and your answer
comes out to be -1

4. r^2 -34 - 70 = 0
r^2 - 104 = 0 Collect your like terms
r^2 = 104 Add 104 to both sides
r = sqrt 104 or 2 * sqrt 26 Take the sqrt of both sides

Answer 2

1. x^3 = 49 x
x^3 - 49 x = 0
x(x^2 - 49 ) = 0

x (x-7) (x+7) = 0

x = 0,+7, -7

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2. 5y^2 = 80 (divide both sides by 5)

y^2 = 16

y = +4, -4
===================

3. 3x^2 + 6x +3 = 0

(3x +3) (x + 1) = 0

x = -1
================
4. r^2 -34 -70 = 0

r^2 = 108


r = +/- square root of 108 or +6 (sq rt of 3) and - 6 (square root of 3)

Answer 3

1. x^3 = 49 x
x^3 - 49 x = 0
x(x^2 - 49 ) = 0

x (x-7) (x+7) = 0

x = +7, -7

==================
2. 5y^2 = 80 (divide both sides by 5)

y^2 = 16

y = +4, -4
===================

3. 3x^2 + 6x +3 = 0

(3x +3) (x + 1) = 0

x = -1
================
4. r^2 -34 -70 = 0

r^2 = 108


r = +/- square root of 108 or +6 (sq rt of 3) and - 6 (square root of 3)

Answer 4

1. Divide both sides by "x" and you'll get x^2 = 49. The square root of 49 is 7, so x=7 or -7.

2. Divide both sides by 5 to get y^2=16. The square root of 16 is 4, so y=4 or -4.

3. Divide both sides by 3 to get x^2+2x+1=0. Using what you know about quadratics, you can factor out the equation. (x+1)(x+1) = 0. In order for the equation to work, x has to equal -1.

4. r^2-34-70 = r^2-104 = 0
so r^2 = 104 therefore r = 2*root of 26

Answer 5

simple.
the first one is x^3 = 49 x 2
step 1: multiply 49 by 2; 98
step 2: the equation is x^3 = 98, so you will put a cubed root sign on both sides (a radical with a small 3 on the top)
step 3: then you know that the cubed root of x^3 is |x|, so in order to find the cubed root of 98, you simply find factors of 98. After that, when you have 3 factors that are the same number that is out of the radical and any remaining numbers stay in the absolute value sign (| |). in this case, the answer is no solution, or cubed root of 98.
second question, divide both sides by 5 and find the square of 16 (don't forget negatives), 3rd question...ask a teacher it's way more complicated to explain, and 4th you should know. Sorry if I'm no help, I don't know how to make math signs on this thing, and I'm not good at explaining over text. ^^;; either that or I totally messed up

Answer 6

definite i could have faith it. that's how you will circulate approximately simplifying this equation. 5(-3x-2)-(x-3)= -4(4x+5)+a million (-15x-10)-x+3= (-16x-20)+a million -15x-10-x+3= -16x-20+a million -15x-x+16x= -20-3+a million+10 -16x+16x= -23+11 0 =/= -12 Yep, constructive that that's -12 (neg. twelve) . a million) Multiply 5 via each and everything interior the parentheses. circulate around the (=) sign and multiply -4 via each and everything in that parentheses. 2) clean the parentheses 3) it quite is the equation devoid of () 4) team like numbers and signs and warning signs (x's with x's) (numbers w/ #) 5) Simplify answer 0 isn't equivalent to a -12

Answer 7

1.)
x^3 = 49x
x^3 - 49x = 0
x(x^2 - 49) = 0
x(x - 7)(x + 7) = 0
x = 0, 7, -7

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2.)
5y^2 = 80
y^2 = 16
y = 4 or -4

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3.)
3x^2 + 6x + 3 = 0
3(x^2 + 2x + 1) = 0
3(x + 1)(x + 1) = 0
x = -1

Answer 8

with a calculator

Answer 9

these questions are so easy they are WHACK

1.7,-7
2.4,-4
3.-1
4.6,-6