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yarga [219]
2 years ago
13

(4, 7); y = 3x + 6 plz help me

Mathematics
1 answer:
nata0808 [166]2 years ago
6 0

y=3x+6.

y=3x-5

I did it with a parallel line

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50 POINTS 10 QUESTIONS! <br><br> Solve each quadratic equation using factoring<br><br> PLEASE HELP!!
Tanzania [10]

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>

6 0
2 years ago
Read 2 more answers
A) Find the average rate of change of the area of a circle with respect to its radius r as r changes from 4 to each of the follo
nata0808 [166]
So the question wants to calculate the change of the area among the circles in respect to its radius changes and the best answer would be the following:
A. 50.27 to 78.54
B.50.27 to 63.62
C.50.27 to 52. 81

The rate of change are the following:
A. 56.24%
B. 26.56%
C. 5.05%
7 0
2 years ago
Read 2 more answers
The dimensions of a right rectangular prism are shown below. 1 in. 1a in. 2 in. in. How many cubes with side lengths of inch are
Dafna1 [17]

First we neeed to find the volume of the rectangular prism

V=\text{ l}\times w\times hV=1\times1\frac{1}{6}\times2\frac{1}{6}=1\times\frac{7}{6}\times\frac{13}{6}=\frac{91}{36}in^3

Volume of cube with side lengths 1/6 = 1/6 x 1/6 x 1/6 = 1/216

The number of cubes needed to fill the prism without gaps or overlaps =

=\frac{91}{36}\text{  divide by }\frac{1}{216}=\frac{91}{36}\times\frac{216}{1}=\frac{19656}{36}=546

3 0
1 year ago
What is the percent decrease from $150 to $100?
GaryK [48]
The best thing to do us to find the difference between $150 and $100. The difference is $50. You've then got to divide $150/$50 and thus gives you 3.333, therefore, the percentage decrease is 33% Hope this helps :)
7 0
3 years ago
Read 2 more answers
Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0
Sonbull [250]

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: x^4+9x^2-4x^2-36

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: x^2(x^2+9)-4(x^2+9)

Now you can rewrite this as (x^2-4)(x^2+9) , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is a^2-b^2=(a+b)(a-b) . Applying that here, we have (x+2)(x-2)(x^2+9) . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: 2x^2+10x-x-5

Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside: 2x(x+5)-1(x+5)

Now you can rewrite the equation as (2x-1)(x+5) . 2x^2 + 9x - 5 is completely factored.

<h3><u>Putting it all together, your factored expression is (x+2)(x-2)(x^2+9)(2x-1)(x+5)=0</u></h3>
5 0
3 years ago
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