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Anit [1.1K]
3 years ago
11

Please need help thank you

Mathematics
1 answer:
valentinak56 [21]3 years ago
5 0
Might be 1? Because it is 100 each
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HELP ASAP
galben [10]

I don’t have a very detailed explanation, however the answer should be “A 90-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left”.

You can clearly see a 90° angle here, it’s obvious. Also, the polygon switch counterclockwise. so, you can see.

hope I helped ^^

3 0
2 years ago
Read 2 more answers
$10.40 to $13.12 is simplify your answer
Bumek [7]

The required simplified value of $10.40 to $13.12 is $2.72.

Given that,
To determine the simplified value of $10.40 to $13.12.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
In the question, asked about the difference between the numbers
$10.40 to $13.12
= 13.12 - 10.40
= $2.72

Thus, the required simplified value of $10.40 to $13.12 is $2.72.

Learn more about simplification here:
brainly.com/question/12501526

#SPJ1

4 0
1 year ago
What is( -10a^5-2a^5+14a^5)(12a^5-6a-10a^3)<br><br> using distributive and the area model.
Delvig [45]
(-10a^5-2a^5+14a^5)(12a^5-6a-10a^3) \\ = 2a^5 (12a^5-6a-10a^3) \\ = 24a^{10} - 12a^6 - 20a^8
8 0
3 years ago
Evaluate function at the given value using the remainder theorem.
garik1379 [7]

Answer:

6

Step-by-step explanation:

f(a)=a^4-2a^3 -10a^2 -7a \\\\f(-2) = (-2)^4-2(-2)^3 -10(-2)^2 -7(-2)\\\\~~~~~~~~~=16-2(-8)-10(4)+14\\\\~~~~~~~~~=30+16-40\\\\~~~~~~~~~=46-40\\\\~~~~~~~~~=6

5 0
1 year ago
What are the vertical and horizontal asymptotes for the function f(x)=<br> 3x2/x2-4
Alecsey [184]

Answer:  f(x) will have vertical asymptotes at x=-2 and x=2 and horizontal asymptote at y=3.

Step-by-step explanation:

Given function: f(x)=\dfrac{3x^2}{x^2-4}

The vertical asymptote occurs for those values of x which make function indeterminate or denominator 0.

i.e. x^2-4=0\Rightarrow\ x^2=4\Rightarrow\ x=\pm2

Hence, f(x) will have vertical asymptotes at x=-2 and x=2.

To find the horizontal asymptote , we can see that the degree of numerator and denominator is same i.e. 2.

So, the graph will horizontal asymptote at y=\dfrac{\text{Coefficient of }x^2\text{ in numerator}}{\text{Coefficient of }x^2\text{ in denominator}}

i.e. y=\dfrac{3}{1}=3

Hence, f(x) will have horizontal asymptote at y=3.

3 0
3 years ago
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