Draw a line plot for the fallowing data measured in inches

Find the missing symbol: 1/6 ? 1/2 < > = please help

Amanda [17]

Answer:

1/6 < 1/2

Step-by-step explanation:

The larger the denominator, the smaller that fraction would be. Both of the numerators are equal, so we don't have to worry about that. Since 6 is larger than 2, we know that 1/2 is larger than 1/6. hope I helped:))

8 0

3 years ago

Read 2 more answers

Solve for the missing arc measure, angle measure, or variable.

zzz [600]

Answer:

m<C I believe is 25 degrees

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

2. Inflation is at a rate of 7% per year. Evan's favorite bread now costs $1.79. What did it cost 10 years ago? How long

zavuch27 [327]

Answer:

It cost $0.91 10 years ago.

It takes 10.24 years for the cost of bread to double.

Step-by-step explanation:

The equation for the price of bread after t years has the following format:

$P(t) = P(0)(1+r)^{t}$

In which P(0) is the current price, and r is the inflation rate, as a decimal.

If we want to find the price for example, 10 years ago, we find P(-10).

Inflation is at a rate of 7% per year. Evan's favorite bread now costs $1.79.

This means that $r = 0.07, P(0) = 1.79$ . So

$P(t) = P(0)(1+r)^{t}$

$P(t) = 1.79(1+0.07)^{t}$

$P(t) = 1.79(1.07)^{t}$

What did it cost 10 years ago?

$P(-10) = 1.79(1.07)^{-10} = 0.91$

It cost $0.91 10 years ago.

How long before the cost of the bread doubles?

This is t for which P(t) = 2P(0) = 2*1.79. So

$P(t) = 1.79(1.07)^{t}$

$2*1.79 = 1.79(1.07)^{t}$

$(1.07)^{t} = 2$

$\log{(1.07)^{t}} = \log{2}$

$t\log{1.07} = \log{2}$

$t = \frac{\log{2}}{\log{1.07}}$

$t = 10.24$

It takes 10.24 years for the cost of bread to double.

7 0

3 years ago

If s(x) = 2 - x2 and t(x) = 3x, which value is equivalent to (st)(-7)

antoniya [11.8K]

Answer:

1029

Step-by-step explanation:

s(-7)=2-49=-47, t(-7)=-21

(st)(-7)=21*49=1029

6 0

3 years ago

The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from th

bazaltina [42]

$\bf \qquad \textit{ inverse proportional variation}\\\\\\
\begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}

&&y=\cfrac{{{ k}}}{x}
\end{array}$

"The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth"

namely "w" varies inversely to $\bf d^2$ or $\bf w=\cfrac{k}{d^2}$

then
"A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. "
namely when w = 192, d = 3900

so $\bf w=\cfrac{k}{d^2}\qquad 
\begin{cases}
w=192\\
d=3900
\end{cases}\implies 192=\cfrac{k}{3900^2}
\\\\\\
\textit{solve for "k", to find the}\\
\textit{"constant of variation"}\\
\textit{once you've found it, plug it back in at }w=\cfrac{k}{d^2}$

7 0

4 years ago