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ololo11 [35]
2 years ago
10

50 POINTS 10 QUESTIONS! Solve each quadratic equation using factoring PLEASE HELP!!

Mathematics
2 answers:
Tanzania [10]2 years ago
6 0

Answer:

Quadratic Equation | Factoring

Solve each quadratic equation using factoring.

1) v² + 5v + 6 = 0

doing middle term factorisation

v²+(3+2)v+6=0

v²+3v+2v+6=0

v(v+3)+2(v+3)=0

(v+3)(v+2)=0

either

<u>v=-3</u>

<u>or</u>

<u>v=-2</u>

2) g² - 3g = 4

keeping all terms in one side

g²-3g-4=0

doing middle term factorisation

g²-(4-1)g-4=0

g²-4g+g-4=0

g(g-4)+1(g-4)=0

(g-4)(g+1)=0

either

<u>g=4</u>

<u>or</u>

<u>g=-1</u>

3)w² + 4w = 0

w(w+4)=0

either

<u>w=0</u>

<u>or</u>

<u>w=-4</u>

4) s² - 8s + 12 = 0

doing middle term factorisation

s²-(6+2)+12=0

s²-6s-2s+12=0

s(s-6)-2(s-6)=0

(s-6)(s-2)=0

either

<u>s=6</u>

<u>or</u>

<u>s=2</u>

5) x ²+ 2x - 35 = 0

doing middle term factorisation

x²+(7-5)x-35=0

x²+7x-5x-35=0

x(x+7)-5(x+7)=0

(x+7)(x-5)=0

either

<u>x=-7</u>

<u>or</u>

<u>x=5</u>

6) r(r + 2) = 99

opening bracket

r²+2r=99

keeping all terms in one side

r²+2r-99=0

r²+(11-9)r-99=0

r²+11r-9r-99=0

r(r+11)-9(r+11)=0

(r+11)(r-9)=0

either

<u>r=-11</u>

<u>or</u>

<u>r=9</u>

7)k(k-4)=-3

opening bracket

k²-4k=-3

keeping all terms in one side

k²-4k+3=0

k²-(3+1)k+3=0

k²-3k-k+3=0

k(k-3)-1(k-3)=0

(k-3)(k-1)=0

either

k=3

or

k=1

8)t²+ 3t + 2 = 0

doing middle term factorisation

t²+(2+1)t+2=0

t²+2t+t+2=0

t(t+2)+1(t+2)=0

(t+2)(t+1)=0

either

<u>t</u><u>=</u><u>-</u><u>2</u>

<u>or</u>

<u>t</u><u>=</u><u>-</u><u>1</u>

9)m ^ 2 - 81 = 0

m²=81

doing square root in both side

\sqrt{m²}=\sqrt{9²}

<u>m=±9</u>

<u>either</u>

<u>m</u><u>=</u><u>9</u>

<u>or</u>

<u>m</u><u>=</u><u>-</u><u>9</u>

10) h²- 17h + 70 = 0

doing middle term factorisation

h²-(10+7)h+70=0

h²-10h-7h+70=0

h(h-10)-7(h-10)=0

(h-10)(h-7)=0

either

<u>h</u><u>=</u><u>1</u><u>0</u>

<u>or</u>

<u>h</u><u>=</u><u>7</u>

slava [35]2 years ago
3 0

Answer:

9) m^2 - 81 = 0

solution,

here,

m^2 - 81 = 0

or, m^2 = 81

or, m^2 = 9^2

or, m = 9 ( removing or cutting both square)

Therefore; m =9

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In parallelogram HJKL if m∡HJK=110∘ find m∡JKL.
pav-90 [236]

Answer:

∠ JKL = 70°

Step-by-step explanation:

Consecutive angles in a parallelogram are supplementary , then

∠ JKL + ∠ HJK = 180°

∠ JKL + 110° = 180° ( subtract 110° from both sides )

∠ JKL = 70°

7 0
3 years ago
FAST, I'LL GIVVE YOU ALOT OF POINTS
Yakvenalex [24]
The RF is 9. You could go to Easycalculations.com/statics/cumulative/relativefrequency to double check :)
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Given the statement, LEI=BNZ and the diagrams shown, determine the value of x.
WARRIOR [948]

Answer:

x = 20

Step-by-step explanation:

Given the triangles are congruent then corresponding angles are congruent, so

∠ N = ∠ E

To calculate ∠ E subtract the sum of the 2 given angles from 180

∠ E = 180 - (x - 5 + 4x - 10)

     = 180 - (5x - 15)

Equate ∠ N and ∠ E

5x - 5 = 180 - 5x + 15 ( add 5x to both sides )

10x - 5 = 195 ( add 5 to both sides )

10x = 200 ( divide both sides by 10 )

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5 0
3 years ago
What is a correct first step in solving the inequality –4(3 – 5x)≥ –6x + 9?
Maksim231197 [3]
Hello,
 the first step should be to distribue:
-4(3-5x)= -12+20x

Teh resolution may be:

-4(3-5x)>=-6x+9
==>-12+20x>=-6x+9
==>20x+6x>=9+12
==>26x>21
==>x>=21/26

But an other way may be used:

-4(3-5x)>=-6x+9
==>3-5x<= -6x/(-4)+9/(-4)
==>-5x-3/2 x<=-9/4 -3

==>-13/2 x <=-21/4

==>x>= -21/4 *(-2/13)
==>x>=21/(2*13)
==>x>=21/26

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I = {2.666666667, 1.166666667
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3 years ago
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