What is the answer to this question?

Arturiano [62]

Very complex, you should ask each individual question, or give more points
general strategy:
get 1 side to equal 0
factor
set each factor to 0
solve

1. factor out the x and set each to zero
x(x+7)=0
x=0

x+7=0
subtract 7
x=-7
x=0 or -7

2. factor out the 6s
6s(s+3)=0
6s=0
s=0

s+3=0
subtract 3
s=-3
s=-3 or 0

3. I'm assuming that the equation was supposed to be t^2+8t+16 so
find what 2 numbers multiply to get 16 and add to get 8
the numbers are 4 and 4
factor
(x+4)(x+4)=0
x+4=0
subtract 4
x=-4

4. find what 2 numbers multipy to get 25 and add to get -10
the numbers are -5 and -5
factor
(x-5)(x-5)=0
set each to zero
x-5=0
add 5
x=5

5. subtract 16 from both sides
h^2+6h-16=0
find what 2 numbers multiply to get -16 and add to get 6
the numbers are -6 and 10
factor
(h-6)(h+10)=0
set each to zero
h-6=0
add 6
h=6

h+10=0
subtract 10
h=-10
h=-10 or 6

6.subtract 5x from both sides
x^2-5x-14=0
find what 2 numbers multiply to get -14 and add to get -5
the numbers are -7 and 2
factor
(x-7)(x+2)=0
set each to zero
x-7=0
add 7
x=7

x+2=0
subtract 2
x=-2
x=-2 or 7

7.add 2r^2 to both sides
2r^2+11r+15=0
difficult so multiply 2 and 15 and get 30
not factor and find what 2 numbers multiply to get 30 and add to get 11
the numbers are 5 nd 6
split 11 up into 5 and 6
2r^2+6r+5r+15=0
group and factor
(2r^2+6r)+(5r+15)=0
factor
2r(r+3)+5(r+3)=0
reverse distribute
(2r+5)(r+3)=0
set each to zero
2r+5=0
subtract 5
2r=-5
divide 2
r=-5/2

r+3=0
subtract 3
r=-3
r=-3 or -5/2

8. this is a difference of 2 perfect squares in which if you have
x^2-y^2 that factors out of (x-y)(x+y) so
x^2-25=0
x^2-5^2=0
(x-5)(x+5)=0
set each to zero
x-5=0
add 5
x=5

x+5=0
subtract 5
x=-5
x=-5 or 5

9. differnce of 2 perfect squares
81-4x^2=0
9^2-(2x)^2=0
(9-2x)(9+2x)=0
set each to zero
9-2x=0
add 2x to both sides
9=2x
divde 2
9/2=x

9+2x=0
subtract 9 from both sidees
2x=-9
divide 2
x=-9/2
x=-9/2 or 9/2

10. subtract 12s from both sides
4s^2-12s+9=0
multiply 4 and 9 and get 36
find what 2 numbers multiply to get 36 and add to get -12
the numbers are -6 and -6
split up the -12s into -6s and -6s
4s^2-6s-6s+9=0
group
(4s^2-6s)+(-6s+9)=0
factor
2s(2s-3)+(-3)(2s-3)=0
reverse distribute
(2s-3)(2s-3)=0
set each to zero
2s-3=0
add 3
2s=3
divide 2
s=3/2

ANWERS:
1. x=0 or -7
2. s=-3 or 0
3. x=-4
4. x=5
5. h=-10 or 6
6. x=-2 or 7
7. r=-3 or -5/2
8. x=-5 or 5
9. x=-9/2 or 9/2
10. s=3/2

4 0

3 years ago

What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$

7 0

3 years ago

How do I solve for X?

lidiya [134]

We use the theory of Pythagoras because the triangle is right.

$AB {}^{2} = AC {}^{2} + BC {}^{2}$
since AB = AC = x
$8 {}^{2} = x {}^{2} + x {}^{2}$
$64 = 2x {}^{2}$
$32 = x {}^{2}$
$x = \sqrt{32}$
$x = \sqrt{16 } \times \sqrt{2}$
$x = 4 \sqrt{2}$

6 0

3 years ago

Read 2 more answers

Help please!! don’t know what to do?!

arlik [135]

Answer:

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Convert 1/5 to a percentage.<br> And<br> Convert 12/100 to a decimal. <br> Please help.

S_A_V [24]