Ask question

Ask question

All categories

2 years ago

8

Suzie made a placemat shaped like an equilateral triangle and wants to put trim around the edges. The length of one side of the

placemat is 18 in.
How much trim does Suzie need?

1 answer:

MAXImum [283]2 years ago

8 0

Suzie needs enough trim to cover the perimeter of the triangle. So she need enough to cover each side of 18. Thus 18 x 3 = 54 and 54 is your answer.

You might be interested in

What is the averge of 70 and 93?

pychu [463]

Add 70+93, then divide by the number of numbers, which is 2.
70+90=163
163/2=81.5
Hope this helps!

3 0

2 years ago

Read 2 more answers

What is the slope if you have (2,1) and (4,4)?<br><br><br>*Ineed this quickly as possible!-Thx

Schach [20]

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{4}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{2}}}\implies \cfrac{3}{2}$

7 0

1 year ago

Read 2 more answers

14.603 rounded to the nearest hundred is

stira [4]

If you mean to the nearest HUNDREDTH the answer is 14.60.

3 0

2 years ago

Read 2 more answers

Please help<br><br>solve this but don't spam<br>

vfiekz [6]

9x-8=9x-3/2 X3
9x-8=27x-9/2
2(9x-2)=27x-9
18x-4 =27x-9
18x-27x=-9+4
-9x =-5
9x=5
X=5/9 answer

4 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago