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3 years ago

10

Alana is making circular picture frames. She needs to put a ribbon along the border of each one. Use what you know about pi and

equivalent ratios to find the circumference of each frame.

1 answer:

ki77a [65]3 years ago

7 0

Idk
sorry
sorrysorrysorrysorry
sorrysorrysorrysorry
sorry

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18 = -5b - 17<br><br><br> help?

cricket20 [7]

Its b= -0.2 or b= 1/-5 whichever one works best

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3 years ago

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Please help me !!!!!!!!!

lidiya [134]

Answer:

f(-2) = 0

f(1) = 12

f(2) = 5

Step-by-step explanation:

f(-2) will lie in the function $f(x) =x^2 - 4$

$\therefore f( - 2) = {( - 2)}^{2} - 4 = 4 - 4 = 0$

f(1) will lie in the function $f(x) =2x^2 +10$

$\therefore f(1) = 2 {(1)}^{2} + 10 = 2 + 10 = 12$

f(2) will lie in the function $f(x) =5$

$\therefore f(2) = 5$

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3 years ago

Linear Equations, Write an equation to represent the plan and describe each variable or number in your equation represent in the

IrinaVladis [17]

Answer:

Step-by-step explanation:

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2 years ago

ln(y+1)-lny=1+3lnx . Express y in terms of x, in a form not involving logarithms. (4 marks, Maths CIE A level)

Delicious77 [7]

$remember$
$ln(x)=log_e(x)$
$and$
$log_x(a)-log_x(b)= log_x( \frac{a}{b} )$
$and$
$log_a(b)=c$ means $a^c=b$
$and$
$blog(a)=log(a^b)$

$so$

$ln(y+1)-ln(y)=1+3ln(x)$
$ln( \frac{y+1}{y} )=1+ln(x^3)$

$minus$ $ln(x^3)$ $from$ $both$ $sides$

$ln( \frac{y+1}{y}-ln(x^3) )=1$
$ln( \frac{y+1}{yx^3})=1$
$log_e( \frac{y+1}{yx^3})=1$

$translate$

$e^1=\frac{y+1}{yx^3}$
$e=\frac{y+1}{yx^3}$

$times$ $both$ $sides$ $by$ $yx^3$

$eyx^3=y+1$

$minus$ $y$ $from$ $both$ $sides$

$eyx^3-y=1$
$y(ex^3-1)=1$

$divide$ $both$ $sides$ $by$ $(ex³-1)$

$y= \frac{1}{ex^3-1}$

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3 years ago

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HELP ME WITH THE EQUATION!!! IF ITS CORRECT ILL GIVE BRAINIST

Andrei [34K]

Answer:

The equation would be F = C - 3m

Step-by-step explanation:

Based on the chart, the fee for each movie you rent costs 3 dollars. The F is the monthly membership fee, the m is the number of movies rented, the 3 is the fee for each movie rented, & the C is the cost.

I really hope that it helps.

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2 years ago