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3 years ago

6

Nellie is buying bird seed to put in her bird feeders. She uses about 10 pounds of bird seed each week, but has room to store an

y extra that she purchases.The bird seed comes in 3 different sizes as given below: 3.5 pounds for $2.52 7.5 pounds for $5.40 11 pounds for $7.92

1 answer:

ololo11 [35]3 years ago

3 0

11 pound for 7.92 is a better deal

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Which rule describes the x-coordinates in the translation below?. 4 2 h 1 M * 3 us A.x+0

Serggg [28]

Answer:

3 = 6 h M

Step-by-step explanation:

4 = 2 h ⋅ ( 1 M )⋅ 3 = 6 h M

you answered one of my questions before now i answer yours lol.

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3 years ago

What is the range of the function represented by the graph?<br> {0,1,2,3} {1,2,3,4} 1

Minchanka [31]

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number 3

Step-by-step explanation:

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3 years ago

Show how to solve the problem 378 x 6 using place value with regrouping. Explain how you knew when to regroup.

KIM [24]

<u>Answer</u>

2,268

<u>Explanation</u>

<u> </u>

<u>By grouping 378 ca</u>n be written as,

378 = 300 + 70 + 8

6× 378 = 6 × (300 + 70 + 8)

= (6×300) + (6×70) + (6×8)

= 1800 + 420 + 48

= 2,268

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2 years ago

Elizabeth is going to roll a fair 6-sided die 600 times. Complete the following statement with the best prediction. Elizabeth wi

SCORPION-xisa [38]

Answer:

Close to 300 but not exaclty 300.

Step-by-step explanation:

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2 years ago

Read 2 more answers

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago