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Genrish500 [490]
3 years ago
6

The graph shows a reflection. True or false: The pre-image and the image are not congruent.

Mathematics
1 answer:
LuckyWell [14K]3 years ago
8 0

Answer:

False

Step-by-step explanation:

That is false

If the pre-image is reflected over any axis, the image will still be congruent to it

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liberstina [14]

it's either a or b, not sure but I think it's b

5 0
3 years ago
Given the function h(x)=x^2-7x+6h(x)=x
Goryan [66]

Answer:

0 (no change)

Step-by-step explanation:

Recall that the average rate of change of a function f(x) over the interval [a,b] is equal to \frac{f(b)-f(a)}{b-a}:

\frac{h(b)-h(a)}{b-a}\\ \\\frac{h(6)-h(1)}{6-1}\\\\\frac{0-0}{5}\\ \\0

Hence, the average rate of change of the function h(x) over the interval [1,6] is 0, or no change.

3 0
2 years ago
What is 4/5 - 2/3?<br> Someone please answer.<br> I have to write up to 20 characters so yh.
disa [49]
Change 4/5 - 2/3 so that the denominator is th same. Let’s try 15.
So, 12/15 - 10/15.
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8 0
3 years ago
Read 2 more answers
A friend has a 83% average before the final exam for a course. That score includes everything but the final, which counts for 25
Klio2033 [76]

Answer:

\mathrm{Best\:course\:grade\:possible:\:}87.25\%,\\\mathrm{Minimum\:score\:on\:final\:to\:earn\:at\:least\:a\:75\%\:for\:the\:course:\:}51\%

Step-by-step explanation:

Assuming the maximum score for the final is 100\%, we can multiply each score by its respective course weight and add them together to give a final score. If your friend did receive this maximum score of 100\%, their overall grade for the course would be:

83(1-0.25)+100(0.25)=\fbox{$87.25\%$}.

To find the minimum score they need to earn a 75% for the course, we set up the following equation:

83(1-0.25)+x(0.25)=75, where x is the minimum score she needs.

Solving, we get:

62.25+x(0.25)=75,\\x(0.25)=12.75,\\x=\fbox{$51\%$}.

8 0
3 years ago
PLEASE HELP VERY URGENT!!!!!!!
Ierofanga [76]

Answer:

The correct option is;

C. (1.6, 1.3)

Step-by-step explanation:

Given that at x = 1.5 the y-values of both equations are y = 1.5 and y = 1 respectively

The x-value > The y-value

The difference in the y-values = 1.5 - 1 = 0.5

At x = 1.6 the y-values of both equations are y = 1.2 and y = 1.4 respectively

The x-value > The y-value

The difference in the y-values = 1.2 - 1.4 = -0.2

At x = 1.7 the y-values of both equations are y = 0.9 and y = 1.8 respectively

The x-value > The first y-value and the x-value < the second y-value

The difference in the y-values = 0.9 - 1.8 = 0.9

Therefore, the approximate y-value can be found by taking the average of both y-values when x = 1.6 where the difference in the y-values is least as follows;

Average y-value at x = 1.6 = (1.2 + 1.4)/2 = 1.3

Therefore, the best approximation of the exact solution is (1.6, 1.3)

By calculation, we have;

-3·x + 6 = 4·x - 5

∴ 7·x = 11

x = 11/7 ≈ 1.57

y = 4 × 11/7 - 5 ≈ 1.29

The solution is (1.57, 1.29)

8 0
3 years ago
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