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3 years ago

Work out V144 - (16 -5 2)2 I

Mathematics

1 answer:

ANTONII [103]3 years ago

3 0

Answer:

-24

Step-by-step explanation:

$\sqrt{144}-(16-5\times 2)^2=\sqrt{144}-(16-10)^2=\sqrt{144}-6^2\\\\=12-36=\boxed{-24}$

Your calculator can be helpful for this.

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△ABC is a right isosceles triangle, the centers of two arcs are the midpoint of AB

mel-nik [20]

Answer:

Step-by-step explanation:

Let's define 3 areas:

S = area of semicircle with radius 6 in (diameter AB)
T = area of quarter circle with radius 6√2 in (radius AC)
U = area of triangle ABC (side lengths 6√2)

The white space between the "moon" and the triangle has area ...

white = T - U

Then the area of the "moon" shape is ...

moon = S -white = S -(T -U) = S -T +U

The area we're asked to find is ...

moon - triangle = (S -T +U) -U = S -T

The formula for the area of a circle of radius r is ...

A = πr²

So, ...

S = (1/2)π(6 in)² = 18π in²

and

T = (1/4)π(6√2 in)² = 18π in²

The difference in areas is S -T = (18π in²) -(18π in²) = 0.

There is no difference between the areas of the "moon" and the triangle.

4 0

3 years ago

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
\\\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
-----------------------------\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
\\\\\\$

$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
\implies 
tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

6 0

4 years ago

Write the decimal equivalent for each rational number. Use a bar over any repeating digits. 2/3 , 3/11 and 8 4/9

kkurt [141]

2/3 = 0.6666666...... (I dont know how to put a bar above the 6)

3/11 = 0.27 (with a bar above the 27

8.4 (with a bar above the 4)

6 0

4 years ago

H + 732 = -194 <br> Solve for H

lilavasa [31]

H + 732 = -194

* combine like terms

H = -194 - 732

H = -926

4 0

3 years ago

Reflect 4, 1 in the x-axis

Marysya12 [62]

Hello there,

When you reflect a point over the x-axis the x-coordinate stays the same while the y-coordinate is its opposite.

In the given point (4,1) when it is reflected over the x-axis you would get (4, -1).

Hope I helped,

Amna

8 0

3 years ago