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Nikolay [14]
3 years ago
5

Evaluate the function below for f(3) f (x) = 3x2 - 5 what’s the answer

Mathematics
1 answer:
Novay_Z [31]3 years ago
4 0
The answer is 13 because you would plug in 3(3) which is 9 then you multiply (9)(2) which is 18 they you subtract 18-5 which equals 13
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Solve for the unknown variable 2/5(3x-4)=-4
Elanso [62]

Answer:

x = -2

Step-by-step explanation:

Solve for x:

(2 (3 x - 4))/5 = -4

Multiply both sides of (2 (3 x - 4))/5 = -4 by 5/2:

(5×2 (3 x - 4))/(2×5) = -4×5/2

5/2×2/5 = (5×2)/(2×5):

(5×2)/(2×5) (3 x - 4) = -4×5/2

5/2 (-4) = (5 (-4))/2:

(5×2 (3 x - 4))/(2×5) = (-4×5)/2

(5×2 (3 x - 4))/(2×5) = (2×5)/(2×5)×(3 x - 4) = 3 x - 4:

3 x - 4 = (-4×5)/2

(-4)/2 = (2 (-2))/2 = -2:

3 x - 4 = 5×-2

5 (-2) = -10:

3 x - 4 = -10

Add 4 to both sides:

3 x + (4 - 4) = 4 - 10

4 - 4 = 0:

3 x = 4 - 10

4 - 10 = -6:

3 x = -6

Divide both sides of 3 x = -6 by 3:

(3 x)/3 = (-6)/3

3/3 = 1:

x = (-6)/3

The gcd of -6 and 3 is 3, so (-6)/3 = (3 (-2))/(3×1) = 3/3×-2 = -2:

Answer:  x = -2

6 0
3 years ago
Read 2 more answers
Solve the inequality.
marysya [2.9K]

Answer:

\huge x >  \frac{7}{3}  \\

Step-by-step explanation:

8( \frac{1}{2} x -  \frac{1}{4} ) > 12 - 2x \\

<u>Expand the terms in the bracket</u>

That's

8( \frac{1}{2} x) - 8( \frac{1}{4} ) > 12 - 2x \\ 4x - 2 > 12 - 2x

Move -2x to the other side of the inequality

4x + 2x - 2 > 12 \\ 6x - 2  > 12

<u>Move - 2 to the other side of the inequality</u>

6x > 12 + 2 \\ 6x > 14

Divide both sides by 6

\frac{6x}{6}  >  \frac{14}{6}  \\

We have the final answer as

x >  \frac{7}{3}  \\

Hope this helps you

5 0
3 years ago
What is the difference between the altitude rule and the leg rule?
aivan3 [116]
The leg rule is hypotenuse/leg and the altitude rule is side 1/altitude=altitude/side 2
6 0
2 years ago
Write a linear function f with f(-3) = 1 and f(13) = 5
GalinKa [24]
D = (5-1)/(13-(-3)) = 0.25

f(13) = 13d + x = 13 * 0.25 + a = 3.25 + a = 5
a = 1.75

f(x) = 1.75 + 0.25x
7 0
3 years ago
How to solve a quadratic equation that does not factor?
svlad2 [7]
Use the formula or complete the square.
The zeroes of the quadratic can be real and rational; real and irrational; complex conjugates.
If the quadratic is ax²+bx+c, x=(-b+√b²-4ac)/2a.
If b² > 4ac the solutions are real. If b²-4ac is a perfect square, the solutions are real and rational; otherwise they’re real but irrational.
If b² < 4ac the solutions are complex.
7 0
3 years ago
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