Mathematics: On two examinations, you have grades of 83 and 89. there is an optional final examination, which counts as one grade. you decide to take the final in order to get a course grade of a, meaning a final average of at least 90

Area of second circle = 56cm² + Area of second circle = 2179.71cm²Step-by-step explanation:let the radius of the small circle be r and that of the big circle be RArea small circle = Area Big circle = Area Big circle - Area small circle = 56cm²- =56cm²The second circle is the bigger circle.Area of second circle - =56cm²Area of second circle = 56cm² + r =26cmArea of second circle = 56cm² + Area of second circle = 56cm² + Area of second circle = 56cm² + Area of second circle = 56cm² + 2123.71Area of...

On two examinations, you have grades of 83 and

Ответ:

thecool498

13.03.2022

Mathematics

Let's say you get x on the third examination.
The average of all three examinations must be at least 90 to get an A.

$\dfrac{x + 83 + 89}{3} \ge 90$

$x + 172 \ge 270$

$x \ge 98$

In order to get an A, you must get at least a 98 on the third examination.

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Ответ:

aime005

14.11.2021

Mathematics

Area of second circle = 56cm² + $\pi r^{2}$

Area of second circle = 2179.71cm²

Step-by-step explanation:

let the radius of the small circle be r and that of the big circle be R

Area small circle = $\pi r^{2}$

Area Big circle = $\pi R^{2}$

Area Big circle - Area small circle = 56cm²

$\pi R^{2}$ - $\pi r^{2}$ =56cm²

The second circle is the bigger circle.

Area of second circle - $\pi r^{2}$ =56cm²

Area of second circle = 56cm² + $\pi r^{2}$

r =26cm

Area of second circle = 56cm² + $\pi r^{2}$

Area of second circle = 56cm² + $\pi* 26^{2}$

Area of second circle = 56cm² + 2123.71

Area of second circle = 2179.71cm²

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On two examinations, you have grades of 83 and 89. there is an optional final examination, which counts as one grade. you decide to take the final in order to get a course grade of a, meaning a final average of at least 90

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