(3n^2-5n+6) + (-8n^2-3n-2)=

A box without a top is made from a rectangular piece of cardboard, with dimensions 27 centimeters by 18 centimeters, by cutting

GaryK [48]

By cutting each corner we have that the resulting dimensions are:
27 - 2x
18 - 2x
Height = x
Therefore, the volume of the box in terms of the variable x, is given by:
V (x) = (x) * (27-2x) * (18-2x)
Answer:
The volume of the box in terms of x is:
(27-x) (18-x) x

8 0

3 years ago

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Can I get some help ???

Nikitich [7]

Answer is D. 288 . i am glad to help

4 0

2 years ago

On Monday, the temperature at 10 a.m. at Sam’s house was –6° Fahrenheit. The

Alika [10]

Answer:

8

Step-by-step explanation:

to find the answer, subtract the second amount (2) from the first amount (-6), -6 - 2 = -8, the temperat decreased by 8

6 0

2 years ago

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A manufacturer has designed a process to produce pipes that are 10 feet long. The distribution of the pipe length, however, is a

zhenek [66]

Answer:

Step-by-step explanation:

8 0

3 years ago

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago