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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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Multiply -8 by -4, then add fifteen! Your answer will be 47!
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3 years ago
Which statement is true ?
8090 [49]

Answer:

f(x)=4x^2

f(2)=4(2)^2

f(2)=4(4)

f(2)=16

2. g(x)=4*2^x

g(2)=4*2^2

g(2)=4*4

g(2)=16

f(2)=g(2)

Both is equal to 16

I hope this help you :)

Step-by-step explanation:

8 0
3 years ago
3. Theresa has 20 coins that are dimes and quarters. Together
KiRa [710]

Answer:

  • 15 dimes
  • 5 quarters

Step-by-step explanation:

Let q represent the number of quarters Theresa has. Then the value of her coins is ...

  0.25q + 0.10(20 -q) = 2.75

  0.15q = 0.75 . . . . . . . . . subtract 2.00, collect terms

  q = 0.75/0.15 = 5

Theresa has 5 quarters and 15 dimes.

3 0
3 years ago
Y= x^2 -5 solve for x
dexar [7]

Y= x^2 -5

We need to solve for x, we need to get x alone

Y= x^2 -5

Lets start by removing -5

Add 5 on both sides

y + 5= x^2 -5 + 5

y + 5= x^2

Now to isolate x , we need to remove the square from x

To remove square , take square root on both sides

+-\sqrt{y+5} = \sqrt{x^2}

square and square root will get cancelled

+-\sqrt{y+5} = x

So x = +\sqrt{y+5} and  x = -\sqrt{y+5}


7 0
3 years ago
Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minim
andreev551 [17]

Answer:

p_v =P(t_{74}    

If we compare the p value and a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

Step-by-step explanation:

Data given and notation    

\bar X=18.81 represent the average lateral recumbency for the sample    

s=8.4 represent the sample standard deviation    

n=75 sample size    

\mu_o =20 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a left tailed  test.  

What are H0 and Ha for this study?    

Null hypothesis:  \mu \geq 20  

Alternative hypothesis :\mu < 20  

Compute the test statistic  

The statistic for this case is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227

The degrees of freedom are given by:

df=n-1=75-1=74    

Give the appropriate conclusion for the test  

Since is a one side left tailed test the p value would be:    

p_v =P(t_{74}    

Conclusion    

If we compare the p value and a significance level for example \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.

3 0
3 years ago
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