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Find the area of a circle with a radius of 7.

Mashutka [201]

Step-by-step explanation:

area of circle= πr²

3.14×7×7

153.86 is the required area

6 0

3 years ago

Help. Find the value of x to the nearest tenth

vodka [1.7K]

I will assume that line 4 bisect line x exactly in half. This means that if we solve the last side of the triangle like seen in the pic below we will know what half of line x is

To find the length of the last side of the triangle (let's call this y) you must use Pythagorean theorem

$a^{2} +b^{2}=c^{2}$

a and b are the legs (the sides that form a perpendicular/right angle)

c is the hypotenuse (the side opposite the right angle)

In this case...

a = 4

b = y

c = 6

^^^Plug these numbers/variables into the theorem

$4^{2} +y^{2} =6^{2}$

solve for y

16 + $y^{2}$ = 36

$y^{2}$ = 20

x = √20

x ≈ 4.472

^^^This is only half of x so if you double it then it will get you the full length of x

4.472 * 2 = 8.944

8.9 (D)

Hope this helped!

~Just a girl in love with Shawn Mendes

6 0

3 years ago

Who could help me out with this

Daniel [21]

A rotation followed by a reflection I think

4 0

3 years ago

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use the numbers 8, 6, and 2 to create one operation to write an expression that includes one exponent and has a value of 8. use

riadik2000 [5.3K]

Answer:

Answer:

Step-by-step explanation:

1) One solution is to write, as our exponent:

2) Because this is special case of the Exponents Law, valid for every base ≠ 0.

3) Hence, including an exponent the numbers 6, 2 whose value is eight.

Therefore our expression is true

Step-by-step explanation:

4 0

3 years ago

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Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago