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algol [13]
3 years ago
5

What is the volume of a cube with length 4 inches? V=lwh

Mathematics
1 answer:
Nata [24]3 years ago
5 0
The answer is:  "64 cubic inches" ;  or, write as:  "64 in³ " .
_______________________________________________________
Explanation:

V = L * w * h ; 

  = (4 inches) * (4 inches) * (4 inches) ;  

(since a "cube" has the same length, width, and height)
  
 =  (4*4*4) inches * inches * inches) ;

 = 16 * 4 * inches³ ;

= 64 inches³ ; or, write as: "64 in³ " ; or, write as:  "64 cubic inches" .
________________________________________________________
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A triangle has side lengths of 5a +3 inches and 2a +3 inches. If the perimeter of the triangle is 3a + 12 inches, which expressi
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The sum of two numbers is fourteen. One number is two less than three times the other. Find the numbers.
Margarita [4]
So, for the first variable, you would use x, and for the second, y. Then, you would write out 2-(3y)=14. once this is done, youd simplify to 2-3y=14. Subtract 2 from both sides to get 3y=12. Then, divide both sides by 3. this will leave you with y=4. From this you can see the other number is ten. Hope this helped! and i apologize if it is incorrect!
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3 years ago
Which best explains whether a triangle with side lengths 2 in., 5 in., and 4 in. is an acute triangle?
Leviafan [203]

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 The triangle is not acute because 2² + 4² < 5².

Step-by-step explanation:

4 0
2 years ago
(1/1+sintheta)=sec^2theta-secthetatantheta pls help me verify this
Xelga [282]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta

Simplify. The denominator uses the difference of two squares pattern:

\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta

Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:

\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta

Split into two separate fractions:

\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta

Rewrite the two fractions:

\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=}  \sec^2\theta - \sec\theta\tan\theta

Hence verified.

8 0
3 years ago
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