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3 years ago

Please help please help please help please help please help please help please help please help

Mathematics

1 answer:

Fudgin [204]3 years ago

8 0

Answer:

two

Step-by-step explanation:

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If f(x)=x²+3x+5, what is f(3+h)?

Simora [160]

You have to substitute 3+h for x, then work out the equation.

(3+h)^2 + 3(3+h) + 5 = 9 + 6h + h^2 + 9 + 3h + 5 =

h^2 + 9h + 23

Answer B.

8 0

3 years ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

7 0

3 years ago

PROBLEM. 11.1: Rewrite These! Rewrite each quotient as a sum or a difference.

Jet001 [13]

Answer:

1. (4x-10)/2

4x/2 - 10/2

2x - 5

2. (1-50x)/-2

1/-2 -50x/-2

25x - 1/2

3. 5(x + 10)/25

(5x + 50)/25

5x/25 + 50/25

1/5 x + 2

4.(-0.2x + 5)/2

-0.2x/2 + 5/2

-0.1x + 2.5

3 0

3 years ago

Find the measure of each angle indicated round to nearest tenth

ludmilkaskok [199]

Is this for a online final if it is I can’t help

3 0

3 years ago

What is 5/9 times 32?

DIA [1.3K]

5/9* 32
= (5*32)/9
= 160/9
= (153+ 7)/9
= 153/9+ 7/9
= 17+ 7/9
= 17 7/9

The final answer is 17 7/9~

7 0

3 years ago