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3 years ago

please don’t answer if you aren’t going to be serious or answer everything

Mathematics

1 answer:

Radda [10]3 years ago

7 0

Answer:

A. They are both vertical lines.

B. Undefined slope

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Write a linear function from the given values: <br><br> f(2)=-2, f(1)=1<br><br><br> f(4)=1, f(8)=4

kap26 [50]

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have:

$f(2)=-2\to(2,\ -2)\\f(1)=1\to(1,\ 1)$ .

Substitute:

$m=\dfrac{1-(-2)}{1-2}=\dfrac{3}{-1}=-3$

Therefore we have $y=-3x+b$ .

Substitute the coordinates of the point (1, 1) to the equation:

$1=-3(1)+b$

$1=-3+b$ <em>add 3 to both sides</em>

$4=b\to b=4$

<h3>Answer: y = -3x + 4</h3>

We have:

$f(4)=1\to(4,\ 1)\\f(8)=4\to(8,\ 4)$

Substitute:

$m=\dfrac{4-1}{8-4}=\dfrac{3}{4}$

Therefore we have $y=\dfrac{3}{4}x+b$ .

Substitute the coordinates of the point (4, 1) to the equation:

$1=\dfrac{3}{4}(4)+b$

$1=3+b$ <em>subtract 3 from both sides</em>

$-2=b\to b=-2$

<h3>Answer: $y=\dfrac{3}{4}x-2$ </h3>

3 0

4 years ago

which of the following properties could be used to rewrite the expression (2/3•1/5)•5/2 as 2/3•(5/2•(5/2•1/5) sorry to ask Su ma

slamgirl [31]

The original expression is given by:
$( \frac{2}{3}*\frac{1}{5})*\frac{5}{2}$
The correct way to rewrite the expression is given by:
$\frac{2}{3}*(\frac{5}{2}*\frac{1}{5})
$
For this, we use two properties:

Associative property:
The way of grouping the factors does not change the result of the multiplication:
$\frac{2}{3}*(\frac{1}{5}*\frac{5}{2})$

Commutative property:
The order of the factors does not vary the product:
$\frac{2}{3}*(\frac{5}{2}*\frac{1}{5})$

8 0

4 years ago

An item is regularly priced at $80 . Sam bought it at a discount of 60% off the regular price. How much did Sam pay?

Nikitich [7]

Sam paid $32 for the item.

6 0

3 years ago

Factor the expression completely.

densk [106]

Step-by-step explanation:

18x²-8

= 2(9x²-4)

= 2(3x+2)(3x-2)

6 0

3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

3 0

1 year ago

*please don’t answer if you aren’t going to be serious or answer everything *

please don’t answer if you aren’t going to be serious or answer everything