X +10=14 <--how does one solve this?

A coat at a store is originally priced at $54.99. The store marks the price down to $32.99. To the nearest percent, what is the

Dmitry_Shevchenko [17]

Answer:

66.6%

Step-by-step explanation:

Step one

given

A coat at a store is originally priced at $54.99

The store marks down the price down to $32.99

Required

the percent decrease

Step two:

the percent decrease= change in price/ old price*100

the percent decrease= 54.99-32.99/54.99*100

the percent decrease= 22/ 32.99*100

the percent decrease= 0.666*100

the percent decrease= 66.6%

6 0

2 years ago

Please help I don't understand what to do

harina [27]

Answer:

This is an a fake answer but I want to supposed to be multiple pictures?? Because you said add an explanation for each picture and is this history? Are you sure it’s math

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Abigail flipped a coin 20 times, and it landed with heads facing up 13 times. What percent of the coin flips landed with heads f

DIA [1.3K]

65% of the coins flipped landed with heads facing up because 13 were faced up out of 20.

Divide:
13/20=0.65=65%

Hope this helped☺☺

5 0

2 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

2 years ago

Read 2 more answers

What is the equation in point-slope form for the line parallel to y=-8x - 5 that contains j(-2, 9)

Julli [10]

Answer:

y - 9 = -8(x + 2)

Step-by-step explanation:

What is the equation in point-slope form for the line parallel to y=-8x - 5 that contains j(-2, 9)

Step 1

y intercept =

y = mx + b

y=-8x - 5

m = -8, b = -5

Point slope form

y - y1 =m( x - x1)

We are given coordinates:

j(-2, 9) = (x1, y1) where ,

x1 = -2

y1 = 9

Hence, our point slope form equation is:

y - 9 = -8(x -(-2))

y - 9 = -8(x + 2)

5 0

2 years ago

X +10=14 <--how does one solve this?​

X +10=14 <--how does one solve this?