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Triss [41]
4 years ago
10

8-6x+11+3x+4x-5=5+3×4

Mathematics
2 answers:
Tanzania [10]4 years ago
3 0

Answer:

x=3

Step-by-step explanation:

x+14=17

-14 -14

x=3

Ivan4 years ago
3 0

Answer:

x = 3

Step-by-step explanation:

Solve for x:

4 x + 3 x - 6 x - 5 + 8 + 11 = 3×4 + 5

3×4 = 12:

4 x + 3 x - 6 x - 5 + 8 + 11 = 12 + 5

Grouping like terms, 4 x + 3 x - 6 x - 5 + 8 + 11 = (-6 x + 3 x + 4 x) + (8 + 11 - 5):

(-6 x + 3 x + 4 x) + (8 + 11 - 5) = 5 + 12

-6 x + 3 x + 4 x = x:

x + (8 + 11 - 5) = 5 + 12

8 + 11 - 5 = 14:

x + 14 = 5 + 12

5 + 12 = 17:

x + 14 = 17

Subtract 14 from both sides:

x + (14 - 14) = 17 - 14

14 - 14 = 0:

x = 17 - 14

17 - 14 = 3:

Answer:  x = 3

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For f(x) = x2 and g(x) = (x − 4)2, in which direction and by how many units should f(x) be shifted to obtain g(x)?
saw5 [17]
Short answer 4 units to the right.

Remark
Graph the two equations.
The purple graph is y = x^2
The black graph is y = (x - 4)^2

Rule
1. When a constant is inside the brackets, the graph moves left or right. To tell which use the second part of the rule.
2. use y = (x + a)^2 as your example.
if a > 0 the graph moves left.
if a < 0 then graph moves right.
Just the opposite of what you would expect.


4 0
3 years ago
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A. SSS Postulate<br> B. SAS Postulate<br> C. Not possible
Novosadov [1.4K]

Answer:

B) SAS

Step-by-step explanation:

there are 2 sides equal and one angle

7 0
3 years ago
Write the sentence as an equation.<br> The sum of 24 and x is - 20.
Firdavs [7]

Answer:

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5 0
3 years ago
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A 10.0 kg and a 2.0 kg cart approach each other on a horizontal frictionless air track. Their
Flura [38]

the  final speed in m/s of the 10.0 kg is 2.53 m/s .

<u>Step-by-step explanation:</u>

Here we have , A 10.0 kg and a 2.0 kg cart approach each other on a horizontal friction less air track. Their  total kinetic energy before collision is 96 ). Assume their collision is elastic. We need to find What is the  final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s . Let's find out:

We know that in an elastic collision :

⇒ Total kinetic energy before collision  = Total kinetic energy after collision

⇒ 96 = \frac{1}{2}M_1(v_1)^2 + \frac{1}{2}M_2(v_2)^2

⇒ 96 = \frac{1}{2}(10)(v_1)^2 + \frac{1}{2}(2)(8)^2

⇒ 96=5(v_1)^2 + 64

⇒ 5(v_1)^2  =32

⇒ (v_1)^2  =6.4

⇒ v_1  =2.53 m/s

Therefore , the  final speed in m/s of the 10.0 kg is 2.53 m/s .

8 0
3 years ago
You move down 5 units and up 5 units. You end at (1,3). Where did you start?
IgorC [24]

Answer:

(1, 3)

Step-by-step explanation:

Since up and down are opposites, the point didn't move at all.

I hope this helps!

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3 years ago
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