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3 years ago

Which rate describes a unit rate of 6:1?

Mathematics

1 answer:

Gemiola [76]3 years ago

4 0

Answer:

6/1

Unit rate = 6

Unit rates are ratios that compare two measurements (units) and the second measurement is 1 (per)

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The following histogram shows the number of items sold at a grocery store at various prices:

Contact [7]

Answer:

C. 2.50, 2.51, 5.00, 5.01, 7.50, 9.00, 10.00

Step-by-step explanation:

8 0

2 years ago

A bakery the price of the cake is nine dollars more than the price of a pie. One day the bakery sold 8 cakes and 14 pies for a t

gavmur [86]

Answer:

The price of the cake is $24 and the price of the Pie is $15

Step-by-step explanation:

Given

<em>Represent price of Cake with C and Price of Pie with P</em>

$C = 9 + P$

Cakes sold = 8

Pies sold = 14

$Total = \$402$

Required

Determine C and P

To represent the cakes and pies sold, we have the following expression

$8C + 14P = 402$

Substitute 9 + P for C

$8(9+P) + 14P = 402$

Open the bracket

$72 + 8P + 14P = 402$

Collect Like Terms

$8P + 14P = 402 - 72$

$22P= 330$

Divide both sides by 22

$P = \frac{330}{22}$

$P = 15$

Recall that

$C = 9 + P$

$C = 9 + 15$

$C = 24$

<em>Hence, the price of the cake is $24 and the price of the Pie is $15</em>

7 0

3 years ago

PLEASE HELP!!!!!!

vlada-n [284]

Answer:

I believe D. B12

5 0

3 years ago

Read 2 more answers

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

Luke and Joshua work at a dry cleaners ironing shirts. Luke can iron 35 shirts per

aniked [119]

Answer:

Step-by-step explanation:

Let x be the number of hours worked by Joshua.

Shirts ironed by Joshua in an hour = 25

Number of shirts ironed by Joshua in x hours = 25x

The number of hours worked by Luke = (x + 3)hours

Shirts ironed by Luke in an hour = 35

No. of shirts ironed by Luke in x hours = 35*(x + 3) = 35x+35*3 = 35x + 105

No. of shirts ironed by Luke and Joshua = 345

25x + 35x +105 = 345

60x = 345 - 105

60x = 240

x = 240/60

x = 4 hours

The number of hours worked by Joshua = 4 hours

The number of hours worked by Luke = 4 +3 = 7 hours

5 0

3 years ago

Read 2 more answers