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beks73 [17]
4 years ago
12

Need help.  collect like terms,    7x²+9x⁴-3x²-x⁴-4x²= _______

Mathematics
2 answers:
alexgriva [62]4 years ago
8 0
7x^2,(-3x^2),(-4x^2) are like terms
9x^4,(-x^4) are like terms
Mashutka [201]4 years ago
3 0
<span>7x²+9x⁴-3x²-x⁴-4x²
(</span>7x²-3x²-4x²) + (9x⁴-x⁴)
(4x²-4x²) + (8x⁴)
(0) + (8x⁴)
8x⁴
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Solve the equations for the given variable. Then place the equations in the table under the correct solution
vekshin1

Placing the equations under the correct solution, we get

x = 3                                      x ≠ 3

-14x = -42                              -5 + x = -9

-\frac{3}{5} +x = \frac{12}{5}                          \frac{x}{3} =9

\frac{x}{4} =\frac{6}{8}                                    

x -5 = -2            

<h3>Solving Linear Equations </h3>

From the question, we are to place the equations under the correct solution

To do this, we will solve the equations one after the other

  • -14x = -42

x = -42/-14

x = 3

  • -5 + x = -9

x = -9 + 5

x = -4

∴ x ≠ 3

  • -\frac{3}{5} +x = \frac{12}{5}

x = \frac{12}{5} +\frac{3}{5}

x = \frac{12+3}{5}

x = \frac{15}{5}

x = 3

  • \frac{x}{4} =\frac{6}{8}

x=\frac{6\times 4}{8}

x=\frac{24}{8}

x = 3

  • \frac{x}{3} =9

x = 3 × 9

x = 27

∴ x ≠ 3

  • x -5 = -2

x = -2 + 5

x = 3

Hence, placing the equations under the correct solution, we get

x = 3                                      x ≠ 3

-14x = -42                              -5 + x = -9

-\frac{3}{5} +x = \frac{12}{5}                          \frac{x}{3} =9

\frac{x}{4} =\frac{6}{8}                                    

x -5 = -2            

Learn more on Solving linear equations here: brainly.com/question/13204213

#SPJ1                  

5 0
2 years ago
The approximate distance between K and L is units. The approximate distance between L and J is units. If you join all three poin
ss7ja [257]

Answer:

The answer is below

Step-by-step explanation:

The distance between two points A(x₁, y₁) and B(x₂, y₂) on the coordinate plane is given by:

AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\

Point J is at (-3, 3), point K is at (4, 3) and point L is at (1, -1). Hence:

The distance between K and L = KL = \sqrt{(-1-3)^2+(1-4)^2} =5\ units

The distance between J and L = JL = \sqrt{(-1-3)^2+(1-(-3))^2} =4\sqrt{2} \ units

The distance between K and J = JK = \sqrt{(3-3)^2+(-3-4)^2} =7\ units

Therefore, the perimeter of triangle JKL is:

Perimeter = KL + JL + JK = 5 + 4√2 + 7 = 17.66 units

8 0
3 years ago
To any of my friends who see this, sorry I haven't been on in a while :(
Tema [17]
The answer is 4 for the missing value
7 0
3 years ago
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Which is the graph of the function f(x) = x2 + 2x – 6? Mark this and return
valina [46]
Vertex: ( -1, -7 )

Focus: ( -1, -27/4 )

Axis of Symmetry: x = -1

Directrix: y = -29/4


X | Y
———
-3| -3
-2| -6
-1 | -7
0 | -6
1 | -3

6 0
3 years ago
Read 2 more answers
Uma makes a scale drawing of a patio. The drawing below shows the two scales she used to plan two patios of different sizes.
Vinil7 [7]

Answer:

Scale 1: 45 m; Scale 2: 60 m

Explanation:

All you do is multiply 15 centimetres by each metre value on each scale to get the above scale.

* Now, you NEVER included expressions, but I did my very best to predict the above imagery.

I am joyous to assist you anytime.

4 0
3 years ago
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