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3 years ago

5

I'm leaving brainly. one final goodbye.

1 answer:

lbvjy [14]3 years ago

8 0

Answer:

awe can i have all of your points??

Step-by-step explanation:

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Which statement is true about the end behavior of the graphed function?

tatyana61 [14]

The <em><u>correct answer</u></em> is:

A) as the x-values go to positive infinity, the functions values go to negative infinity.

Explanation:

We can see in the graph that the right hand portion continues downward to negative infinity. The right hand side of the graph is "as x approaches positive infinity," since x continues to grow larger and larger. This means as x approaches positive infinity, the value of the function approaches negative infinity.

8 0

3 years ago

Read 2 more answers

Help please I’m so dumb for this

Keith_Richards [23]

Answer:

you are dumbbbbbb

Step-by-step explanation:

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3 0

4 years ago

Read 2 more answers

Pls solve question 4b. Verrry urgent . Ok tyy

kodGreya [7K]

Answer:

$y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}$

Step-by-step explanation:

Given equation:

$y = 3x^2 - 9x + 17$

Factor out 3 from the first 2 terms:

$y = 3(x^2 - 3x) + 17$

Divide the coefficient of x by 2 and square it: (-3 ÷ 2)² = 9/4

Add this inside the parentheses and subtract the distributed value of it outside the parentheses:

$y = 3\left(x^2 - 3x+\dfrac{9}{4}\right) + 17-\dfrac{27}{4}$

Factor the parentheses and combine the constants:

$y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}$

8 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

3 years ago

Craig measured a length as 0.034 millimeters. What is the margin of error?

Salsk061 [2.6K]

Answer:

0.0335mm to 0.0345mm

Step-by-step explanation:

7 0

3 years ago