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1 year ago

H(x) = -2x+5 g(x) = 2x² + x Find (h+g)(1)

Mathematics

1 answer:

Westkost [7]1 year ago

4 0

Are you evaluating functions for this? If so, here you go:

Given the functions to simplify (h + g)(1), the answer you seek would be 6 I believe.

Hope this helps!

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From the 1,2,3,4,5 which values make the inequality n+1 > 4 true?

ELEN [110]

Answer:

yes it is

Step-by-step explanation:

hope I helped;)

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2 years ago

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A colleague has been tutoring six students in 11th grade to prepare for the ACT. This colleague has asked you to evaluate the pe

marissa [1.9K]

We have been given that a colleague has been tutoring six students in 11th grade to prepare for the ACT. Student scores were as follows: 20, 18, 16, 15, 23, 20. We are asked to find the mean of the ACT scores.

We will use mean formula to solve our given problem.

$\text{Mean}=\frac{\text{Sum of terms}}{\text{Number of terms}}$

$\text{Mean}=\frac{20+18+16+15+23+20}{6}$

$\text{Mean}=\frac{112}{6}$

$\text{Mean}=18.66666$

Upon rounding to nearest whole number, we will get:

$\text{Mean}\approx 19$

Therefore, the mean of the ACT scores is 19 and option 'c' is the correct choice.

5 0

2 years ago

How do you do this. Plz explain with steps.

Marysya12 [62]

What you are going to do is multiply what you have in parenthesis by two the answer is 2Xsquared and 2Y to the fourth

8 0

2 years ago

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Please help me

Snowcat [4.5K]

Answer:

Option 1, 2, and 4 are correct. Option 3 is incorrect.

6 0

2 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

2 years ago