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tatuchka [14]
3 years ago
14

What is the solution of the equation (x -- 5)2 + 3(x - 5) + 9 = 0? Use u substitution and the quadratic formula to solve

Mathematics
1 answer:
Westkost [7]3 years ago
6 0

Answer:

so what I found is x= 16/5

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Help me PLEASEEEEEEEEEEEEE
Andrej [43]

The correct answer is 2.5 cm

Explanation:

In any polygon, the area refers to the total space occupied by the polygon. In the case of triangles, the area can be calculated by multiplying the base and height and then dividing the by 2. This means b x h / 2 = area of a triangle. According to this, the area of the triangle presented is 15

5 cm (base) x  6 cm (height) = 30 cm

30 cm / 2 =  15 cm (total area)

This also means the area of the rectangle is 15 considering both figures have the same area. Additionally, the area in rectangles is calculated by multiplying the length base by the width. This implies in the case presented 6 cm x w cm = 15 cm (total area), and you can determine the value of w by using a simple equation as

6 x w = 15

w = 15 / 6

w =  2.5

Also, you can know this is the correct value because 6 cm (length) x 2.5 (width) = 15 cm which is the correct area

4 0
3 years ago
Work with a partner. A survey asked 200 people to name their favorite
SSSSS [86.1K]

i dont get it either

8 0
2 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
Solve the equation 16x^3+16x^2-x-1=0 given that -1/4 is a zero of f(x)= 16x^3+16x^2-x-1. Solution set is ___
Keith_Richards [23]

Answer:

I love algebra anyways

The ans is in the picture with the  steps how i got it

(hope this helps can i plz have brainlist :D hehe)

Step-by-step explanation:

8 0
3 years ago
Graph y=-3x +7 and the line that is<br> parallel to it that passes through (4,1).
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Y = -3x + 13

Steps and the graph are on paper

6 0
3 years ago
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