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____ [38]
3 years ago
10

After graphing What is the solution of y= -3x -1 and y= x + 3

Mathematics
1 answer:
Tresset [83]3 years ago
5 0

Answer: (3,4)

Step-by-step explanation:

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Refer to the figure below to complete the following item,
Aloiza [94]

Answer:

50 degrees

Step-by-step explanation:

3 0
3 years ago
Find the inverse of the given function.<br><br> f(x) = -1/2√x + 3, x ≥ -3
inna [77]
<span>Find the inverse of the given function.

f(x) = -1/2√x + 3, x ≥ -3

I will have to assume that you meant f(x) = -(1/2)sqrt(x) + 3.  If you actually meant f(x) = -(1/2)sqrt(x+3), then obviously the correct result would be different.

1.  Replace "f(x)" by "y:"  y </span>= -(1/2)sqrt(x) + 3
 2.  Interchange x and y:  x = -(1/2)sqrt(y) + 3
3.  Solve for y:  x-3=-(1/2)sqrt(y), so that 2(3-x)= sqrt(y) and y=+sqrt(2[3-x])
4.  Replace "y" with 

           -1
        f      (x) = sqrt(2[3-x])

Here, there are restrictions on x, since the domain of the sqrt function does not include - numbers.  The domain here is (-infinity,3]
4 0
3 years ago
Read 2 more answers
Solve the system by substitution. Check your solution.
Zepler [3.9K]

Answer:

a.  (15, 15)

Step-by-step explanation:

We start with those two equations:

1) a - 1.2b = -3

2) 0.2b + 0.6a = 12

We'll begin by modifying equation #1 to isolate a:

a = -3 + 1.2b

Then we'll use this value for a in the second equation:

0.2b + 0.6 (-3 + 1.2b) = 12

0.2b - 1.8 + 0.72b = 12

0.92b = 13.8

b = 15

Then we'll place that value of b in the first equation to find a:

a - 1.2 (15) = -3

a - 18 = -3

a = 15

3 0
3 years ago
The product of a binomial and a trinomial is x 3 + 3 x 2 − x + 2 x 2 + 6 x − 2. Which expression is equivalent to this product a
Amiraneli [1.4K]

The answer is A

x3 + 5x2 + 5x – 2

Brainiest? :)

4 0
3 years ago
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A pair of jeans sells for $49.99. The sales tax rate is 6.5%. What is the total cost for the jeans including sales tax?​ Please
sergij07 [2.7K]

Answer: Total cost would be $53. 24 including tax which is $3.25

Step-by-step explanation:

6 0
3 years ago
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