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Len [333]
2 years ago
14

PLEASE HELP I GIVE BRAINLIEST

Mathematics
1 answer:
RoseWind [281]2 years ago
4 0

can u take a better pic :) so its more clear!!

Step-by-step explanation:

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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
Read 2 more answers
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
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