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Bezzdna [24]
3 years ago
11

Please help.. tysm if you do

Mathematics
1 answer:
iris [78.8K]3 years ago
4 0

Answer:

the answer is option A one graph

and I think or am sure it is correct ans.

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How to use. Rounding when estimating
Lina20 [59]
You just round the numbers
4 0
3 years ago
Let f(x)=2x^2-x-1 and let g(x)=x-1 Which statement is true?​
sergij07 [2.7K]

Answer:

B is true

Step-by-step explanation:

given f(x) = 2x² - x - 1 and g(x) = x - 1

A

If x = - 1 is a root then f(- 1) = 0

f(- 1) = 2(- 1)² - (- 1) - 1 = 2 + 1 - 1 = 2 ≠ 0 ← False

B

If (x - 1) is a factor then x = 1 is a root and f(1) = 0

f(1) = 2(1)² - 1 - 1 = 2 - 1 - 1 = 0

⇒ g(x) = x - 1 is a factor of f(x) ← True

C

If x = 2 is a root then f(2) = 0

f(2) = 2(2)² - 2 - 1 = 8 - 3 = 5 ≠ 0 ← False

D

\frac{x2x^2-x-1}{x-1} = \frac{(2x+1)(x-1)}{x-1}

Cancel the factor (x - 1) on the numerator/ denominator, leaving

\frac{f(x)}{g(x)} = 2x + 1 with remainder 0 ≠ 2 ← False

7 0
4 years ago
Express 19,796 correct to 2 decimal places​
True [87]

Answer:

19,80

Step-by-step explanation:

Hi, you just need to make a rounding: 19.796 => 19.80. Hope it helped!

8 0
3 years ago
Who knows what erattas is
Minchanka [31]

Answer:

Errata is originally the plural of the singular Latin noun erratum. Like many such borrowed nouns ( agenda; candelabra ), it came by the mid-17th century to be used as a singular noun, meaning “a list of errors or corrections to be made (in a book).”

Step-by-step explanation:

7 0
3 years ago
Please help me if you know how to do this!
lukranit [14]

Question 21

Let's complete the square

y = 3x^2 + 6x + 5

y-5 = 3x^2 + 6x

y - 5 = 3(x^2 + 2x)

y - 5 = 3(x^2 + 2x + 1 - 1)

y - 5 = 3(x^2+2x+1) - 3

y - 5 = 3(x+1)^2 - 3

y = 3(x+1)^2 - 3 + 5

y = 3(x+1)^2 + 2

Answer: Choice D

============================================

Question 22

Through trial and error you should find that choice D is the answer

Basically you plug in each of the given answer choices and see which results in a true statement.

For instance, with choice A we have

y < -4(x+1)^2 - 3

-7 < -4(0+1)^2 - 3

-7 < -7

which is false, so we eliminate choice A

Choice D is the answer because

y < -4(x+1)^2 - 3

-9 < -4(-2+1)^2 - 3

-9 < -7

which is true since -9 is to the left of -7 on the number line.

============================================

Question 25

Answer: Choice B

Explanation:

The quantity (x-4)^2 is always positive regardless of what you pick for x. This is because we are squaring the (x-4). Squaring a negative leads to a positive. Eg: (-4)^2 = 16

Adding on a positive to (x-4)^2 makes the result even more positive. Therefore (x-4)^2 + 1 > 0 is true for any real number x.

Visually this means all solutions of y > (x-4)^2 + 1 reside in quadrants 1 and 2, which are above the x axis.

5 0
3 years ago
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