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vampirchik [111]

3 years ago

11

HOPEFULLY MY LAST ONE!!!!

2 answers:

Serhud [2]3 years ago

4 0

Answer:

1/3

Step-by-step explanation:

Rise/run

10/30

I hope this is right and that it helps.

Gnoma [55]3 years ago

4 0

Use these points to find out the slope: (10,60) , (40,70)
Slope: 1/3

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Which value must be added to the expression x^2+x to make it a perfect-square trinomial? A.1/4 B.1/2 C.1 D.4

barxatty [35]

Hello from MrBillDoesMath!

Answer:

Choice A, 1/4

Discussion:

Consider the perfect square

(x + a)^2 = x^2 + (2a)x + a^2

The constant term (a^2) equals 1/2 the coefficient of x (i.e. 2a), squared.

Let's apply this idea to x^2 + x

x^2 + x = => as 2 * 1/2 = 1

x^2 + ( 2 * 1/2)x =

( x^2 + (2* 1/2)x + ( 1/2) ^2 ) - (1/2) ^2 =

as constant term to add is 1/2 coefficient of x (that is, 1/2) and

(1/2)^2 - (1/2)^2 = 0

(x + 1/2) ^ 2 - (1/2)^2

In other words add the constant (1/2)^2 = 1/4, which is Choice A.

Thank you,

MrB

7 0

3 years ago

Evan collected data about the amount of time each student in his math class spends studying. If the range of these data is 2, wh

djyliett [7]

A range of 2 reveals that the smallest and largest data values are 2 units away from each other.

<h3>What does the range reveal?</h3>

Range is the difference between the highest and lowest values of a set of observations. Range is a measure of variation.

Range = highest value - lowest value

To learn more about range, please check: brainly.com/question/12372689

#SPJ1

6 0

2 years ago

If anybody is good at math , can you drop your number or something

Luden [163]

Answer:

its the third one

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Leah wrote 2 different fractions with the same denominator.Both fractions were less than 1. Can their sum equal 1? Can their sum

Alex777 [14]

Answer:

1. Yes

2. Yes

Step-by-step explanation:

Leah wrote 2 different fractions with the same denominator. Both fractions were less than 1.

1. Can their sum equal 1?

Let Leah fractions be $\dfrac{6}{7}$ and $\dfrac{1}{7}.$ Both these fractions have the same denominators and are less than 1. Find their sum:

$\dfrac{6}{7}+\dfrac{1}{7}=\dfrac{7}{7}=1$

2. Can their sum be greater than 1?

Let Leah fractions be $\dfrac{6}{7}$ and $\dfrac{5}{7}$ . Both these fractions have the same denominators and are less than 1. Find their sum:

$\dfrac{6}{7}+\dfrac{5}{7}=\dfrac{11}{7}=1\dfrac{4}{7}>1$

7 0

4 years ago

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r >

Ira Lisetskai [31]

Answer:

The other pairs are:

$(a)\ (2, \frac{5\pi}{6}) \to$ $(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3}) \to$ $(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4}) \to$ $(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

See attachment for plots

Step-by-step explanation:

Given

$(a)\ (2, \frac{5\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

Solving (a): Plot a, b and c

See attachment for plots

Solving (b): Find other pairs for $r > 0$ and $r < 0$

The general rule is that:

The other points can be derived using

$(r, \theta) = (r, \theta + 2n\pi)$

and

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

Let $n =1$ ---- You can assume any value of n

So, we have:

$(r, \theta) = (r, \theta + 2n\pi)$

$(r, \theta) = (r, \theta + 2*1*\pi)$

$(r, \theta) = (r, \theta + 2\pi)$

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

$(r, \theta) = (-r, \theta + (2*1 + 1)\pi)$

$(r, \theta) = (-r, \theta + (2 + 1)\pi)$

$(r, \theta) = (-r, \theta + 3\pi)$

$(a)\ (2, \frac{5\pi}{6})$

$r = 2\ \ \ \ \theta = \frac{5\pi}{6}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi}{6} + 2\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi+12\pi}{6})$

$(2, \frac{5\pi}{6}) = (2, \frac{17\pi}{6})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi}{6} + 3\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi+18\pi}{6})$

$(2, \frac{5\pi}{6}) = (-2, \frac{23\pi}{6})$

The other pairs are:

$(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$r = 1\ \ \ \theta = -\frac{2\pi}{3}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(1, -\frac{2\pi}{3}) = (1, -\frac{2\pi}{3} + 2\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (1, \frac{-2\pi+6\pi}{3})$

$(1, -\frac{2\pi}{3}) = (1, \frac{4\pi}{3})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(1, -\frac{2\pi}{3}) = (-1, -\frac{2\pi}{3} + 3\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (-1, \frac{-2\pi+9\pi}{3})$

$(1, -\frac{2\pi}{3}) = (-1, \frac{7\pi}{3})$

The other pairs are:

$(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

$r = -1 \ \ \ \ \theta = \frac{-5\pi}{4}$

So, the pairs are

$(r, \theta) = (r, \theta + 2\pi)$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi}{4} + 2\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi+8\pi}{4} )$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{3\pi}{4} )$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(-1, \frac{-5\pi}{4}) = (-(-1), \frac{-5\pi}{4}+ 3\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (1, \frac{-5\pi+12\pi}{4})$

$(-1, \frac{-5\pi}{4}) = (1, \frac{7\pi}{4})$

So, the other pairs are:

$(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

5 0

3 years ago