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grin007 [14]
2 years ago
14

IF YOU SOLVE THIS WHOLE PAGE I'LL MARK YOU THE BRAINLIEST​

Mathematics
2 answers:
evablogger [386]2 years ago
5 0

Answer:

so this the answer for your question.but im sorry for questions 14,16 and 17 cause i cant solve.

jekas [21]2 years ago
4 0

Step-by-step explanation:

x=9

x=5

x=-12

x=0

-x=0

x=2

x=11

x=6

x=10

x=8

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What is the area of the rectangle? Round to the nearest square inch.
Mademuasel [1]
Asking the Math Gods...

A=W*L

9*3=27in^2
9*5=45in^2

3 0
2 years ago
The sum of two integers is 81.if the product is 1088.what are the integers
arlik [135]

Answer:

(a;b)={(17; 64); (64; 17)}

Step-by-step explanation:

a+b=81 => b=81-a

a*b=1088

a*(81-a)=1088

-a²+81a=1088

a²-81a+1088=0

a²-64a-17a+1088=0

a(a-64)-17(a-64)=0

(a-17)(a-64)=0

=> a=17 and a=64

for a=17 => b=81-17=64

for a=64 => b=81-64=17

(a;b)={(17; 64); (64; 17)}

3 0
3 years ago
Please HJelp QUICKLY
torisob [31]
<span>Along the x axis:
5 - 3 = 2 blocks

</span>Along the y axis:
8 - 1 = 7 blocks

Total:
2 + 7 = 9 blocks

7 0
2 years ago
Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2
tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

<u>1. 2x^2 - 7x + 3</u>

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

  • a = 2
  • b = -7
  • c = 3

we can use the quadratic formula: x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}

Substitute the a, b, and c values into the quadratic formula: x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}

Now simplify using the laws of pemdas: x=\frac{7\pm\sqrt{(49)-(24)} }{4}

Simplify even further: x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}

Now split this equation into two equations to solve for x: x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

\boxed {x=3,\frac{1}{2} }

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

  • x - 3 = 0
  • 2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

<u>2. 3x^2 + 7x + 2</u>

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

  • 3x + 1 = 0
  • x + 2 = 0

Solve.

<u>First factor:</u> 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

<u>Second factor:</u> x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

\boxed {x = -\frac{1}{3} , -2}

7 0
3 years ago
Select all the expressions that are equivalent to the polynomial below.
katrin2010 [14]
The answers are: D and C and A and E
because: 


B 3(x - 5) - 2(6x2 + 9x + 5)=<span><span><span>−<span>12<span>x^2</span></span></span>−<span>15x</span></span>−<span>25

</span></span>F (4x2 - 13x - 7) - (16x2 + 9x - 5)=<span><span>−<span>12<span>x2</span></span></span>−<span>22x</span></span>−<span>2

E </span>(-15x2 + 9x - 10) + (3x2 - 10x - 5)=<span><span>−<span>12<span>x^2</span></span></span>−x</span>−<span>15</span>
4 0
3 years ago
Read 2 more answers
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