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ololo11 [35]
3 years ago
10

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

Mathematics
2 answers:
Tanzania [10]3 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]3 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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Answer:

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Step-by-step explanation:

Given: the approximate height, h, in meters, travelled by golf balls hit with two different clubs over a horizontal distance of d meters is given by the functions: h=-0.002d^2+0.3d\,,\,h=-0.004d^2+0.5d

To find: distances when the ball is at the same height when either of the clubs  are used. Also, to find this height

Solution:

(a)

-0.002d^2+0.3d=-0.004d^2+0.5d\\d(-0.002d+0.3)=d(-0.004d+0.5)\\-0.002d+0.004d=0.5-0.3\\0.002d=0.2\\d=\frac{0.2}{0.002}=100\,\,metres

(b)

Put d = 100 in h=-0.002d^2+0.3d

h=-0.002(100)^2+0.3(100)\\=-20+30\\=10\,\,metres

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Jennifer wants to take a trip around the world. She plans to deposit $125 at the beginning of each month into an investment with
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The formula of the future value of annuity due is
A=p [(1+r/k)^(kn)-1)/(r/k)]×(1+r/k)
A future value of annuity due
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x=-1/2

Step-by-step explanation:

add 2 to both sides

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subtract x from both sides

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Quick Computing Company produces
olganol [36]

Answer:

c(x) = 2x^2 + 6x + 25

Completed question;

Quick Computing Company produces calculators. They have found that the cost, c(x), of making x calculators is a quadratic function in terms of x. The company also discovered that it costs $45 to produce 2 calculators, $81 to produce 4 calculators, and $285 to produce 10 calculators. Derive the function c(x).

Step-by-step explanation:

Given that;

the cost, c(x), of making x calculators is a quadratic function in terms of x.

c(x) = ax^2 + bx + c

Substituting the 3 case scenarios given;

it costs $45 to produce 2 calculators,

45 = a(2^2) + b(2) + c

45 = 4a + 2b +c .......1

$81 to produce 4 calculators,

81 = a(4^2) + b(4) + c

81 = 16a + 4b + c .......2

and $285 to produce 10 calculators.

285 = a(10^2) + b(10) + c

285 = 100a + 10b + c .......3

Solving the simultaneous equation;

Subtracting equation 1 from 2, we have;

36 = 12a + 2b ......4

Subtracting equation 1 from 3

240 = 96a + 8b .......5

Multiply equation 4 by 4

144 = 48a + 8b ......6

Subtracting equation 6 from 5, we have;

96 = 48a

a = 96/48

a = 2

Substituting a = 2 into equation 4;

36 = 12(2) + 2b

36 = 24 + 2b

2b = 36-24 = 12

b = 12/2 = 6

b = 6

Substituting a and b into equation 1;

45 = 4(2) + 2(6) +c

45 = 8 + 12 + c

c = 45 - (8+12)

c = 25

Since a = 2 , b = 6 and c = 25, the quadratic equation for c(x) is ;

c(x) = ax^2 + bx + c

c(x) = 2x^2 + 6x + 25

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