All categories

3 years ago

Bird seed is four dollars a bag of you buy two bags how much change will you get from ten dollar

Mathematics

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

Step-by-step explanation:

2 $4 bags = $8

$10 - $8 = $2 change

You might be interested in

Solve for x<br><br> (picture is included below)

Anna11 [10]

Solve for x:

4x - 7 = 3

Add 7 to both sides

4x = 10

Divide both sides by 4

x = 10/4

Simplify:

x = 5/2

Answer for question 1: x = 5/2

13 + 2x/3 = 15

Multiply both sides by 3

39 + 2x = 45

Subtract 39 from both sides

2x = 6

Divide both sides by 2

x = 3

Answer for question 2: x = 3

10x + 7 = 15

Subtract 7 from both sides

10x = 8

Divide both sides by 10

x = 8/10

Simplify

x = 4/5

Answer for question 3: x = 4/5

8 0

2 years ago

Read 2 more answers

Solve for y: 1/3y+7≥12<br><br> A) y ≤ 5/3<br> B) y ≥ 15<br> C) y ≤ 5<br> D) y ≥ 57

Klio2033 [76]

Answer:

B) y ≥ 15

Step-by-step explanation:

6 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

How many pairs of parallel sides does a trapezoid have? none/ one/ three/ two

telo118 [61]

A trapezoid has two parallel sides, but the top one is shorter than the bottom one

3 0

2 years ago

Larry is shipping nails that weigh a total of 53 pounds. he must divide the nails equally into 4 boxes. how many nails in each b

lina2011 [118]

13 boxes and there will be 1/4 of a box left over

3 0

3 years ago

Read 2 more answers