All categories

2 years ago

I need help with this

Mathematics

1 answer:

ankoles [38]2 years ago

5 0

Answer:

Step-by-step explanation:

I gotchu fam. The first thing that you would want to do is find the slope. The slope is the change in y divided by the change in x. Based on the points on the line, the change in y is 2 because it goes up two, and the change in x is one because it goes over 1. This would make your slope 2/1, or just 2! Now we have our slope, so we want to find our y-intercept. There are two ways that we can do this. The first way is just eyeballing it, but sometimes that's not always accurate. We're just going to do that because its faster (lol). It looks around -2. With this information, we have our entire equation! y=mx+b. we can plug in our information. y=2x-2. And that's how your answer's A! Hope this helps.

You might be interested in

1. The formula for the volume of a cone is

Andre45 [30]

Answer:

$r=\sqrt{\frac{3V}{\pi\ h}}$

Step-by-step explanation:

The volume of cone is given by:

$V=\frac{1}{3}\pi\ r^2\ h$

where $r$ is the radius of the base and $h$ is the height of the cone.

We need to solve for $r$

we have $V=\frac{1}{3}\ \pi\ r^2\ h$

Multiplying both sides by 3 to remove $\frac{1}{3}$

⇒ $3\times V=3\times \frac{1}{3}\ \pi\ r^2\ h$

⇒ $3V=\pi\ r^2\ h$

dividing both sides by $\pi\ h$

⇒ $\frac{3V}{\pi\ h}=\frac{\pi r^2 h}{\pi\ h}$

⇒ $\frac{3V}{\pi\ h}=r^2$

Taking square root both sides:

⇒ $\sqrt{\frac{3V}{\pi\ h}}=\sqrt{r^2}$

⇒ $\sqrt{\frac{3V}{\pi\ h}}=r$

∴ $r=\sqrt{\frac{3V}{\pi\ h}}$

3 0

3 years ago

Read 2 more answers

Multiple choice plz plz help! Thanks:)

Minchanka [31]

A- Since the radius is 22, the area is 1520.53, or, 1520.5 simplified.

3 0

3 years ago

A circle has a radius of 12. What is the diameter?

AveGali [126]

Answer:

24.

Step-by-step explanation:

the diameter is twice the radius

7 0

2 years ago

Read 2 more answers

Of three positive integers, the second is twice the first, and the third is twice the second. One of these integers is 17 more t

marusya05 [52]

X = 1st integer, y = 2nd integer, z = 3rd integer

y = 2x
z = 2y......z = 2(2x)...z = 4x

y = x + 17
2x = x + 17
2x - x = 17
x = 17

y = 2x
y = 2(17)
y = 34

z = 4x
z = 4(17)
z = 68

and the sum would be : 17 + 34 + 68 = 119

8 0

2 years ago

Drag the expressions into the boxes to correctly complete the table.

lora16 [44]

Answer:

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form $ax^n$ .

Here:

$n$ = non-negative integer

$a$ = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

$x^4+\frac{5}{x^3}-\sqrt{x}+8$

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains $\sqrt{x}$ , so it is not a polynomial.

Also it contains the term $\frac{5}{x^3}$ which can be written as $5x^{-3}$ , meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression $x^4+\frac{5}{x^3}-\sqrt{x}+8$ is not a polynomial.

Given the expression

$-x^5+7x-\frac{1}{2}x^2+9$

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

$\left|x\right|^2+4\sqrt{x}-2$

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

$x^3-4x-3$

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

$\frac{4}{x^2-4x+3}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

3 0

3 years ago