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dsp73
2 years ago
7

Rewrite the following sentence to remove the “not” (make sure to use correct spelling and punctuation)::

English
1 answer:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

She was a good student.

Explanation:

I hope this helps, have a good day!

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Fernando evaluated the expression below. StartFraction 5 (9 minus 5) over 2 EndFraction + (negative 2) (negative 5) + (negative
Juli2301 [7.4K]

Question:

Fernando evaluated the expression below.

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = \frac{5(4)}{2} -10 + 9 = \frac{20}{2} - 10 + 9 = 10 - 10 + 9 = 9

What was Fernando's error?

- Fernando evaluated the numerator of the fraction incorrectly.

- Fernando simplified 20/2 incorrectly.

- Fernando incorrectly found the product of –2 and –5.

- Fernando evaluated (-3)² incorrectly.

Answer:

- Fernando incorrectly found the product of –2 and –5.

Explanation:

Given

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = \frac{5(4)}{2} -10 + 9 = \frac{20}{2} - 10 + 9 = 10 - 10 + 9 = 9

Required

Spot the error

The error in this evaluation is the product of -2 and -5 at step 2

When a negative number (-2) is multiplied by a negative number (-5), the outcome of the multiplication is always a positive number (10).

So, the result of -2 * -5 is 10 but Fernando incorrectly calculated it as -10.

Solving the expression, correctly

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = \frac{5(4)}{2} + 10 + 9

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = \frac{20}{2} + 10 + 9

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = 10 + 10 + 9

\frac{5(9-5)}{2} + (-2)(-5) + (-3)^2 = 29

Hence, the actual result of the expression is 29 (not 9 as calculated by Fernando).

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3 years ago
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Answer: So she could find employment

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In the play, ''Twelfth Night '' by William Shakespeare, Viola was shipwrecked and rescued but was now in a foreign land. She therefore needed a way to support herself and so needed employment.

In order to get over gender barriers to employment, she disguised herself as a man and picked the name Cessario to enter into the service of Duke Orsino, who rules Illyria.

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