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stepladder [879]

7 months ago

8

If nine out of every sixteen students prefer to write with pens over pencils, what proportion of students prefer pens to pencils

?

1 answer:

Roman55 [17]7 months ago

5 0

Answer:

7

Step-by-step explanation:

U put the six aside then take the n subtract the 9 nd take de answer n add it with de 6 u've put away at 1st

Lyk dis

6+(10 - 9)

6+(1)

6+1

=7

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1+1I need help and fast i've been trying to solve this one for 30 minutes now.

Liula [17]

Answer:

2

Step-by-step explanation:

somebody has to answer eventually lol

6 0

2 years ago

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0

2 years ago

Use the distributive property to remove the parentheses.<br> -4(-6y + 2w-4)

olga_2 [115]

Answer: 24y - 8w + 16

Step-by-step explanation:

Basically to remove the parenthesis you just distribute the -4 outside of the parenthesi and you get 24y - 8w + 16.

4 0

2 years ago

Plsss helpppppppppp if you do, you a cool guy or girl

STatiana [176]

9514 1404 393

Answer:

Tyler
2 hundredths of a mile

Step-by-step explanation:

The graph is a little difficult to read, but we note that there are 6 grid lines between times that are 2 minutes apart. So, each grid line stands for 2/6 = 1/3 minute.

At the 1-mile mark, the graph crosses 1 grid line above 8 minutes, indicating it takes Tyler 8 1/3 minutes to run 1 mile.

Then in 10 minutes, Tyler will run ...

distance = speed · time = 1 mile/(8 1/3 min) · 10 min

= 1/(25/3)·10 = 10·3/25 = 30/25 = 1.2 . . . . miles

__

The equation tells you that Elena runs each mile in 8.5 minutes. To see how far she runs in 10 minutes, we can solve ...

10 = 8.5x

x = 10/8.5 ≈ 1.18 . . . . miles

So, Tyler runs farther in 10 minutes by a distance of ...

1.20 -1.18 = 0.02 . . . . miles

8 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

2 years ago