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WINSTONCH [101]
2 years ago
9

Some pleaseeeeee help!! f(x) = 3x + 4 + 2 g(x) = 4 Find (f +g)(x).

Mathematics
1 answer:
Cerrena [4.2K]2 years ago
5 0

Answer:

3x^2 + 10x

Step-by-step explanation:

If u wrote it correctly it would be ((x+4+2)+4)(x)

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A 5 in by 7in photograph is surrounded by a frame of uniform width the area of the frame equals the area of the photograph find
alexira [117]

<span>Let x be the width of the frame. 
Area of the picture is 5*7 = 35 sq in</span>

Area of the frame given as the same, frame = 35 sq in also

<span>
Therefore the total area (picture & frame) will be 70 sq in</span>

<span>

A simple equation would be made and it will look like this: (7+2x) * (5+2x) = 70

</span>

<span>Using the FOIL method
35 + 14x + 10x + 4x^2 = 70
4x^2 + 24x + 35 - 70 = 0; subtract 70 from both sides:
4x^2 + 24x - 35 = 0</span>

<span>
unluckily this will not factor easily, so we should use the quadratic equation:
where: a = 4; b = 24; c = -35</span><span>
</span><span> x = -b sqrt b^2 – 4*a*c divided by 2 * a</span><span>

  x = -24</span> ±<span> sqrt 24^2 -4*4*(-35) divided by 2*4

 x = -24 </span>±<span>sqrt 576 - -(560) divided by 8

 x = -24 </span>± sqrt 576 + 560 divided by 8<span>

 x = -24 </span>± sqrt1136  divided by 8<span>

 x = -24 + 33.7046 divided by 8
x =   9.7046 divided by 8
x =1.213 inches is the answer</span>

<span> </span>

6 0
3 years ago
-5(a-6)+ 2a Help me with this equation pleaseee asap
Ne4ueva [31]

Answer:

-3a+30

Step-by-step explanation:

-5a+30+2a.

-5a+2a=-3a

-3a+30

6 0
2 years ago
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Ethan went on a 6 mile kayaking expedition down the Taos Box on the Rio Grande River last week. The expedition Was divided into
Ksivusya [100]
Chibi form*
HI there. Here is your answer:
First section: 2:30-3:10 {40 min.}
Second secton:3:10-3:30 {20 min.}
Third stage:3:30-4:10 {40 min.}

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4 0
3 years ago
The coordinate plane below represents a city. Points A through F are schools in the city.
STALIN [3.7K]

Part A--

In our graph, point A is at (2, -3) and point E is at (3, 1). Since we want to include these and only these points, the procedure is to first find a line through A and E, then take what's to the right as the TRUE space. What's to the left is the FALSE space.

To find the through A and E, we need the slope, m, first.

m = y₂ - y₁ / x₂- x₁

Choose x₁ as 3, y₁ as 1, x₂ as 2, y₂ as -3.

m = \frac{(-3) - 1 }{2-3}

m = \frac{-4}{-1} = \frac{4}{1} = 4

The slope of the line is 4. Using x = 3, y = 1 and m = 4, we can find the equation of the line and report it in slope-intercept form.

y = mx + b

1 = 3 * 4 + b

1 = 12 + b

-11 = b

With a slope of 4 and y-intercept of -11, the equation of the line is y = 4x -11.

Now we make the inequality. Choose one symbol and a test value. If the tests are true, keep it. If false, switch it. Anything to the right of A and E is what we want to be true; anything to the left is false. Let's choose a point to the right - any will work, but (5,0) will be ours. We want to make the inequality true when x = 5, y = 0.

y ≥ 4x - 11 (use ≥ to include the points)

0 ≥ 20 - 11

0 ≥ 9 FALSE

Try switching the inequality.

y ≤ 4x - 11

0 ≤ 20 -11

0 ≤ 9 TRUE

To check, choose a point to the left (where it's false). (0,0) works well.

y ≤ 4x - 11

0 < 0 -11

0 ≤ -11 FALSE

So the line we want is y ≤ 4x -11

To create a second equation, let's divide the city in two diagonally (like cutting bread). Let's go from bottom left to top right, and the equation for that line is

y = x (slope is 1, y-intercept is 0). Also, you can pick any horizontal or vertical line that works)

Let's use (5.0) as our test point yet again. We want it to true, so we want x to be bigger than y. Flip it around, and y is smaller than x.

A working system for the school choice is:

y ≤ 4x -11

y ≤ x

Part B--

We created the second equation to contain points A and E, but we can test each point in the two inequalities to see if they are true.

Point A is (2, -3) and Point E is (3, 1)

y ≤ x

Testing Point A: -3 ≤ 2 TRUE

Testing Point E: 1 ≤ 3 TRUE

y ≤ 4x -11

Testing Point A:

-3 ≤ 4(2) - 11

-3 ≤ 8 -11

-3 ≤ -3 TRUE

Testing Point E

1 ≤ 4(3) - 11

1 ≤ 12 -11

1 ≤ 1 TRUE

Both points work pairwise in both equations.

Part C--

We are now told that William can only go to schools where the equation y < - x - 1 is TRUE. Let's graph our line. Luckily, the line is already in slope-intercept form. The y-intercept is -1, so (0-1) is on the graph. A slope of 1 means that if we go down one unit and to the right one unit (negative y, positive x), that gives us a second point. So (1, -2) is on the graph. We need to find where the equation is true, so let's test x = 0, y = 0 and see which side to shade.

y < -x - 1

0 < 0 -1

0 < -1 TRUE

Anything above the line will make the equation true. Since it is < and not ≤, the points on the line are not included. Anything above the line is included. You could either (1) shade above the line (but not the line inclusive - use a broken line) and find the schools in the shaded area or (2) test the six points one by one to see if make the inequality y < - x - 1 true or false.

4 0
3 years ago
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Find all seventh roots of unity and sketch them on the axis below.
jarptica [38.1K]
There is no picture??
4 0
3 years ago
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