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3 years ago

Help me with my homework

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

4 0

Answer: A. There are three terms

Concept:

In an algebraic expression, there are several specific terms:

Constant: the individual number that has no variable with it, or in east word, it is isolated.
Variable: any unknown value, which is usually presented by alphabets.
Coefficients: the numbers that are bounded with a variable, usually in front of a variable.
Terms: the number of individuals present in the expression, which is constants and [variable + coefficients].
Like terms: terms that have the same variables and powers.

If you are still confused, you may refer to the attachment below for a graphical explanation.

Solve:

a. There are three terms. $\boxed{FALSE}$

- 14b

- -7

- 8b

- b

There are in total four individuals, which means there are four terms.

b. There is one constant term. $\boxed{TRUE}$

- 14b is a [variable + coefficient]

- -7 is a [constant]

- 8b is a [variable + coefficient]

- b is a [variable]

Therefore, there is one constant term.

c. The coefficients are 14, 8, and 1 $\boxed{TRUE}$

- 14b →→→ 14 is the [coefficient]

- -7 →→→ is a [constant]

- 8b →→→ 8 is the [coefficient]

- b →→→ 1 is the [coefficient]

Therefore, 14, 8, and 1 are coefficients.

d. Three of the terms are like terms $\boxed{TRUE}$

- 14b →→→ b is the [variable]

- -7 →→→ is a [constant]

- 8b →→→ b is the [variable]

- b →→→ b is the [variable]

Since 14b, 8b, and b share the common variable, there are three like terms.

Hope this helps!! :)

Please let me know if you have any questions

EleoNora [17]3 years ago

3 0

Answer:

a because I dont know why but some one told me it was this

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Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla

Ahat [919]

(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

$\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0$

and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

5 0

2 years ago

If you are also able to explain this please do! I'd greatly appreciate it!

pogonyaev

You're given a table of x values and an equation. To solve, plug in the values for x.

First box: y=2(-3)-1=-7
Second box: y=2(-2)-1=-5.

See if you can solve the third one.

7 0

3 years ago

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A rectangular prism has a length of 5 1/8 feet, a width of 7 1/2 feet, and a height of 2 feet. What is the volume of the prism?

maksim [4K]

For this case we have that by definition, the volume of a rectangular prism is given by:

$V = A_ {b} * h$

Where:

$A_ {b}:$ It is the area of the base

h: It's the height

According to the data we have:

$length = 5 \frac {1} {8} = \frac {8 * 5 + 1} {8} = \frac {41} {8}$

$width = 7 \frac {1} {2} = \frac {2 * 7 + 1} {2} = \frac {15} {2}$

Then:

$A_ {b} = \frac {41} {8} * \frac {15} {2} = \frac {615} {16}$

Thus, the volume is:

$V = \frac {615} {16} * 2 = \frac {1230} {16} = 76.875$

Answer:

$76.875 \ ft ^ 3$

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3 years ago

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vekshin1

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