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Natali5045456 [20]

3 years ago

6

The circumference of a circle measures 15.7 centimeters. What is the measure of its diameter?

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

A. 12 cm

there you go

IceJOKER [234]3 years ago

4 0

Answer:

a

Step-by-step explanation:

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Lim. √2x -√3x-a/√x-√a<br>x approaches a

Ludmilka [50]

Answer: $\sqrt{2}a-\sqrt{3}a-2\sqrt{a}$

Step-by-step explanation:

$\lim _{x\to \:a}\left(\sqrt{2}x-\sqrt{3}x-\frac{a}{\sqrt{x}}-\sqrt{a}\right)$

$=\sqrt{2}a-\sqrt{3}a-\frac{a}{\sqrt{a}}-\sqrt{a}$

$=\sqrt{2}a-\sqrt{3}a-2\sqrt{a}$

3 0

3 years ago

Slove for y show all your work 12y + 4 = 8y-12

Vedmedyk [2.9K]

Answer:

y = -4

Step-by-step explanation:

12y + 4 = 8y-12

Subtract 8y from each side

12y - 8y +4 = -12

4y +4 = -12

Subtract 4 from each side

4y +4-4 = -12 -4

4y = -16

Divide by 4

4y/4 = -16/4

y = -4

8 0

3 years ago

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What is the unit price of a quart of milk for $0.89?

astraxan [27]

Answer:

$3.65/gallon APEX

Step-by-step explanation:

7 0

2 years ago

Rewrite in simplest terms: 4(-5u +3) – 4(8u - 1)

viktelen [127]

Answer:

You distribute then combine

your answer is <u><em>-52u+16</em></u>

Step-by-step explanation:

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3 years ago

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago