Answer:
13
Step-by-step explanation:
Replace X with 2
Evaluate the function. g(x) = 3x^2 – 2x + 5 Find f(2)
g(x) = 3(2)^2 – 2(2) + 5
Next conduct PEMDAS
Exponents are first so solve 2^2 which is 2 x 2 = 4
g(x) = 3(4) – 2(2) + 5
Next step is multiplication multiply 3 x 4 and 2x2
g(x) = 12 – 4 + 5
conduct adding and subtracting left from right
g(x) = 13
Answer:
m = 4
you can find the answer for this on m a t h w a y (without the spaces)
I hope this helps!
Answer:
Step-by-step explanation:
To write fractions with a common denominator, you will most likely need to scale some numbers up! I will explain how.
Let's try it with the fractions
2
3
and
3
12
12 is larger than 3, so we will have to multiply the 3 by some number to equal 12. (We are really finding the Least Common Multiple of the two denominators!) To do this, you have to multiply the 3 by 4, because 3x4=12. But now the numerator doesn't match the denominator. When you scale the denominator up, you have to scale the numerator up too! So the 2 must be multiplied by 4 also.
Now you have the following:
8
12
and
3
12
These fractions now have common denominators! Now they're all set for adding or subtracting fractions.
Try another:
2
6
and
3
5
: The least common multiple of 6 and 5 is 30. (the product of the denominators)
Transform each fraction by multiplying by "1":
2
6
⋅
5
5
=
10
30
and
3
5
⋅
6
6
=
18
30
One last problem:
4
9
and
7
6
What is the least common multiple of 9 and 6? Could you use 54? Absolutely, but it is not the LEAST number that you could use. How about 18? YES!
4
9
⋅
2
2
=
8
18
and
7
6
⋅
3
3
=
21
18
Ready to go...
Hope this helped!
Answer:

Step-by-step explanation:
<u>Equation of a circle</u>

(where (a, b) is the center and r is the radius)
Given:
- center = (5, 1)
- diameter = 4√5
Diameter = 2r (where r is the radius)
⇒ r = 4√5 ÷ 2 = 2√5
Substituting these values into the equation:


Step-by-step explanation:
7(4x + 9) - 4
= (7 × 4x) + (7 × 9) - 4
= 28x + 63 - 4
= <u>28x + 59</u>
So, 7(4x + 9) - 4 is equivalent to 28x + 59 (D)
<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>