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3 years ago

Please help - it’s due at 12:00 No links pls

Mathematics

2 answers:

hram777 [196]3 years ago

7 0

Answer: Keep going

Step-by-step explanation:

iren [92.7K]3 years ago

5 0

The answers is here I upload a chicken in the Beth room so you can see him ok thank you very mi

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I need help pleasee. i will give brainlest

Degger [83]

Answer:

Step-by-step explanation:

7 0

2 years ago

Prove that: 1+cot²a / 1 + cosec a = cosec a

fgiga [73]

Step-by-step explanation:

LHS= 1 + cot^2A/1+cosecA

=(1+cosecA + cot^2A)/1+cosecA

{By using basic trigonometric theorem cot^2A = cosec^2A - 1}

=(1+cosecA + cosec^2A - 1)/(1+cosecA)

=(cosecA + cosec^2A)/(1+cosecA)

=[cosecA (1+cosecA)]/(1+cosecA)

{Cancelling 1+cosecA}

=cosecA

3 0

3 years ago

Henry knows that the circumference of a circle is 18 pie inches. What is the area of the circle?

Aneli [31]

Answer:

254.34

Step-by-step explanation:

This is because A=C2

4π which is 18x18 divided 4x 3.14 which equals 254.34

6 0

3 years ago

Read 2 more answers

Simplify radicals problems attached please help!

Finger [1]

Answer:

G7. (2√3)/3

G8. -2+√7

G9. (6 +2√2 -3√3 -√6)/7

Step-by-step explanation:

G7.

$\dfrac{2}{\sqrt{3}}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$

G8.

$\dfrac{3}{2 +\sqrt{7}}=\dfrac{3}{2+\sqrt{7}}\cdot\dfrac{2-\sqrt{7}}{2-\sqrt{7}}=\dfrac{6-3\sqrt{7}}{2^2-(\sqrt{7})^2}\\\\=\dfrac{6-3\sqrt{7}}{-3}=\sqrt{7}-2$

G9.

$\dfrac{2-\sqrt{3}}{3-\sqrt{2}}=\dfrac{2-\sqrt{3}}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}=\dfrac{(2-\sqrt{3})(3+\sqrt{2})}{9-2}\\\\=\dfrac{6+2\sqrt{2}-3\sqrt{3}-\sqrt{6}}{7}$

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Comment on the problems

In most cases, these expressions are the simplest possible (take the least amount of ink to draw, and take the fewest math operations to evaluate). What seems to be intended is that the denominator be made a rational number. This is done by multiplying the given fraction by a fraction equal to 1 that has the same denominator but with the sign of the radical reversed (unless, as in the first case, the radical is by itself).

The purpose of doing this is to take advantage of the fact that (a-b)(a+b) = a²-b², so if "a" or "b" is a square root, that root will not be seen in the product. In problem G9, we see this can make the numerator quite messy--not exactly a simpler form--but all the irrational numbers are in the numerator.

8 0

4 years ago

Find the slope please

zavuch27 [327]

Hi student, let me help you out! :)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We are asked to find the slope of this graph; we are provided with two points:

$\star~\mathrm{(6,0)}$