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Brilliant_brown [7]
2 years ago
7

Simplify 7 sqare root 3 takeaway 2 sqare root 3 add sqaure root 3 takeaway 3 sqaure root 3

Mathematics
1 answer:
Paha777 [63]2 years ago
8 0

Answer:

tel:+18446450594

Step-by-step explanation:

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pentagon [3]

Answer:

answer is b juicy how to do it

8 0
3 years ago
Given the functions f(x) = 2x -1 and g(x) =
Novosadov [1.4K]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

  • f(x) = 2x - 1
  • g(x) =  2 - x

a)

f(x) + g(x) = (2x - 1) + (2 - x)

                = 2x -1 + 2 - x

                = x + 1

b)

f(x) - g(x) = (2x - 1) - (2 - x)

              = 2x - 1 - 2 + x

               = 3x - 3

c)

g(-5) - f(-5)

Putting x = -5 in g(x) = 2 - x

g(x) = 2 - x

g(-5) = 2 - (-5) = 2+5 = 7

Putting x = -5 in f(x) = 2x - 1

f(x) = 2x - 1

f(-5) = 2(-5) - 1

       = -10 - 1

        = -11

Thus,

g(-5) - f(-5) = 7 - (-11) = 7+11 = 18

d)

f(x).g(x) = (2x - 1) (2 - x) = -2x² + 5x - 2

e)

f(g(x)) = f(2-x)

          = 2(2-x)-1

          = 4-2x-1

          = 3-2x

6 0
3 years ago
Determine the inequality signs on a graph
TiliK225 [7]

Answer:

y = x^2

theres nothing shaded so its not an inequality.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Please help! alg 1! I really need a explaintion fastttt
Gelneren [198K]

Answer:

  1. y = -5x +10
  2. (0, 10)
  3. (1, 5)
  4. see attached

Step-by-step explanation:

1. Subtract 5x from both sides to put the equation in slope-intercept form:

  y = -5x +10

__

2. The y-intercept is the point corresponding to x=0. The y-value when x=0 is the constant in the equation: 10. Then the point is ...

  (x, y) = (0, 10)

You may notice this is one of the points listed in part 4, and is also used in question 3.

__

3. The x-value computed is 1; the y-value computed is 5. The point is ...

  (x, y) = (1, 5)

You may notice this is one of the points listed in part 4.

__

4. See attached

7 0
3 years ago
Help please and thank you!
Usimov [2.4K]

Answer:

Option A

Explanation:

\rightarrow \sf sin\left(\dfrac{7\pi }{6}\right)

\rightarrow \sf \sin \left(\pi +\dfrac{\pi }{6}\right)

\rightarrow \sin \left(\pi \right)\cos \left(\dfrac{\pi }{6}\right)+\cos \left(\pi \right)\sin \left(\dfrac{\pi }{6}\right)

\rightarrow \sf 0 *  \dfrac{\sqrt{3}}{2}+\left(-1\right) * \dfrac{1}{2}

\rightarrow \sf -\dfrac{1}{2}

7 0
2 years ago
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