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Elodia [21]
3 years ago
6

Pls help this is due today ! Find the value x:

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

Answer:

x= 12

Step-by-step explanation:

180° = (3x - 6) + (8x + 4) + 50°  because all interior angles of a triangle add up to 180°

180° = 11x +48°

11x = 132°

x = 12

You might be interested in
explain the benefits of each of the three forms of quadratic equations, standard form, vertex form, and factored form. What do t
mamaluj [8]

Answer:

Summary:

Standard form allow us to quickly find the y-intercept.

Vertex form allow us to quickly locate the vertex.

And factored form allows us to quickly determine the roots/zeros.

Step-by-step explanation:

The three forms of quadratics are the standard form, vertex form, and the factored form. Each of them reveals a specific part about the quadratic.

Standard Form:

The standard form of a quadratic is:

ax^2+bx+c

There are only two details that can be conveyed by a quadratic in standard form immediately: (1) the leading coefficient a, and (2) the y-intercept.

The leading coefficient a will tell us if the parabola curves upwards or downwards.

And the constant c will give us the y-intercept.

Vertex Form:

The vertex form of a quadratic is:

a(x-h)^2+k

There are also two details that can be conveyed by a quadratic in vertex form:  (1) the vertex, and (2) the leading coefficient.

The leading coefficient is given by a. Again, this tells us the orientation of the parabola.

And the vertex is given by (h, k).

Hence, in my opinion, vertex form is the best form of a quadratic since it immediately reveals the vertex, the most important aspect of a quadratic.

Factored Form:

The factored form of a quadratic is:

a(x-p)(x-q)

Where p and q are the zeros/roots/solutions of the quadratic.

Again, factored form gives us two details about the quadratic: (1) the leading coefficient, and (2) the zeros.

The zeros tells us when the parabola crosses the x-axis, which can assist in graphing.

Summary:

Therefore, each form of a quadratic equation has its own benefits.

Standard form allow us to find the y-intercept.

Vertex form allow us to quickly locate the vertex.

And factored form allows us to quickly determine the roots/zeros.

Hence, depending on the question, each form can be useful in its own way.

4 0
3 years ago
Consider this system of linear equations: 22x − y = -23 (equation 1) 11x + 12y = 1 (equation 2) To find the value of x, you need
loris [4]
For the first equation just solve for y
then plug into the 2nd one
then solve for x
6 0
3 years ago
I really don’t understand what the questions asking? How do I solve this
Eddi Din [679]

Answer:

$17.85

Step-by-step explanation:

Let the dollar amount per shirt be represented by s.

Let the dollar amount per pants be represented by p.

You have the following system:

3s+2p=104.81

2s+1p=61.33

If I multiply the bottom equation by -2, I can set the system up for elimination since the equations are already and the same form and I will have 2p and -2p in the same column there.  2p+(-2p)=0

Multiplying equation 2 by -2:

3s+2p=104.81

-4s-2p=-122.66

-----------------------Add the equations:

-1s      =-17.85

Multiply both sides by -1:

s=17.85

So one shirt cost $17.85

5 0
3 years ago
......................
kaheart [24]
.......................
5 0
3 years ago
A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val
Taya2010 [7]

Answer:

The local maximum and minimum values are:

Local maximum

g(0) = 0

Local minima

g(5.118) = -350.90

g(-1.368) = -9.90

Step-by-step explanation:

Let be g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}. The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

g(x) = x^{2}\cdot (x^{2}-5\cdot x -14)

Then, first and second derivatives of the function are, respectively:

First derivative

g'(x) = 2\cdot x \cdot (x^{2}-5\cdot x -14) + x^{2}\cdot (2\cdot x -5)

g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}

g'(x) = 4\cdot x^{3}-15\cdot x^{2}-28\cdot x

g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)

Second derivative

g''(x) = 12\cdot x^{2}-30\cdot x -28

Now, let equalize the first derivative to solve and solve the resulting equation:

x\cdot (4\cdot x^{2}-15\cdot x -28) = 0

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

x\cdot (x-5.118)\cdot (x+1.368) = 0

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28

g''(0) = -28

This value leads to a local maximum.

x = 5.118

g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28

g''(5.118) = 132.787

This value leads to a local minimum.

x = -1.368

g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28

g''(-1.368) = 35.497

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}

g(0) = 0

Local minima

g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}

g(5.118) = -350.90

g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}

g(-1.368) = -9.90

7 0
4 years ago
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