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Semenov [28]
4 years ago
6

For the linear function f(x) = 7x - 4, find the range of f(x) at x = -2,0, and 2.

Mathematics
1 answer:
kenny6666 [7]4 years ago
8 0

Answer:

(-2, -18) and (2, 10)

Step-by-step explanation:

the answer is -18 and 10

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What is the product (4x)( -3x^6)(-7x^3)
11111nata11111 [884]

Answer:

  84x^10

Step-by-step explanation:

The coefficients multiply, and the exponents add.

  (4x)( -3x^6)(-7x^3) = (4)(-3)(-7)x^(1+6+3) = 84x^10

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The rule for exponents is ...

  (x^a)(x^b) = x^(a+b)

7 0
3 years ago
Pls help i will give 5 starts and brainliest question is down below 9:22
OleMash [197]

4) -0.3 = -1/3 < -1/7 since the bigger a negative number is the less positive it is. 1/8 is positive so it is clear that it should be placed last. & thus do all options this way.

5) B

Area of a triangle = ½ bh.

where b = base &

h = height.

Area of outer shape = (½ x 16 x 12) in^2.

= 96 in^2.

Area of unshaded inner shape = (½ x 3 x 4) in^2.

= 6 in^2.

Area of shaded region = (96 – 6) in^2.

= 90 in^2.

4 0
2 years ago
This week, the art museum gave away 1,200 tickets to the Greco exhibit, which was 150 percent as many tickets as it gave away la
svet-max [94.6K]

Answer:

Step 1 is ya answer!!! UwU

Step-by-step explanation:

Sorry if im wrong!!! T>T But it should be right!!! :D

4 0
3 years ago
What is<br> -1 5/6 - (-3 1/4) ?
lions [1.4K]

Answer:

1\frac{5}{12}

Step-by-step explanation:

Eliminating a negative and changing our operation

1\frac{5}{6} +3\frac{1}{4}

Rewriting our equation with parts separated

-1-\frac{5}{6} +3+\frac{1}{4}

Solving the whole number parts

-1+3=2

Solving the fraction parts

-\frac{5}{6} +\frac{1}{4} =[?]

Find the LCD of 5/6 and 1/4 and rewrite to solve with the equivalent fractions.

LCD = 12

-\frac{10}{12} +\frac{3}{12}=-\frac{7}{12}

Combining the whole and fraction parts

2-\frac{7}{12} =1\frac{5}{12}

[RevyBreeze]

8 0
3 years ago
Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5
mariarad [96]
The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).
5 0
3 years ago
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