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MArishka [77]
3 years ago
5

Solve by graphing y= -4x+5 y=3x-2

Mathematics
1 answer:
Hatshy [7]3 years ago
4 0
Y = -4x + 5
y = 3x - 2

  -4x + 5 = 3x - 2
+ 4x       + 4x
           5 = 7x - 2
        + 2        + 2
           7 = 7x
           7     7
           1 = x
           y = -4x + 5
           y = -4(1) + 5
           y = -4 + 5
           y = 1
     (x, y) = (1, 1)
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Half a number increased by 15 is equal to the sum of 5 and the product of 3 and the number.
son4ous [18]

Answer:

(x/2) + 15 = 5 + 3x

Step-by-step explanation:

Follow the steps in the equation. X is your number. First you halve it, then you add 15. Next, set it equal to the other side. The second side is 5 + (because it is the sum) and 3x (the product of 3 and x means multiply them)

3 0
3 years ago
Polynomial equation when zeros are -4, -2, 1, 4
jek_recluse [69]
<h2>The required poynomial equation is x^{4}+x^{3}-18x^{2}-16x+32.</h2>

Step-by-step explanation:

Given,

The zeros of the polynomial are - 4, - 2, 1 and 4

To find, the polynomial equation = ?

We know that,

The polynomial equation = (x - A)(x - B)(x - C)(x - D)

= (x + 4)(x + 2)(x - 1)(x - 4)

= [(x + 4)(x - 4)] [(x + 2)(x - 1)]

= (x^{2}-16)(x^2-x+2x-2)

=  (x^{2}-16)(x^2+x-2)

= x^{4}+x^{3}-2x^{2}-16x^{2}-16x+32

= x^{4}+x^{3}-18x^{2}-16x+32

Thus, the required poynomial equation is x^{4}+x^{3}-18x^{2}-16x+32.

4 0
3 years ago
Given a triangle MTN, prove that
cupoosta [38]

Answer:

<em></em>\angle m + \angle t + \angle n = 180<em></em>

<em></em>

Step-by-step explanation:

Required

Show that:

\angle m + \angle t + \angle n = 180^o

To make the proof easier, I've added a screenshot of the triangle.

We make use of alternate angles to complete the proof.

In the attached triangle, the two angles beside \angle m are alternate to \angle t and \angle n

i.e.

\angle 1 = \angle t

\angle 2 = \angle n

Using angle on a straight line theorem, we have:

\angle 1 + \angle m + \angle 2 = 180

Substitute values for (1) and (2)

\angle t + \angle m + \angle n = 180

Rewrite as:

<em></em>\angle m + \angle t + \angle n = 180<em> -- proved</em>

5 0
3 years ago
6^7/6^5 in index form<br>help me
brilliants [131]

Answer:

12^12

Step-by-step explanation:

can u please help me

7 0
3 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
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