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2 years ago

7

Dr.Imos plants a vegetable garden each year.He plants one fourth of his garden in corn.and one half of his garden in tomatoes,He

buys 240 seeds.If he pays $0.05 per seed for all vegetables he plants, how much does he spend on seeds? use S.O.L.V.E

1 answer:

12345 [234]2 years ago

3 0

Answer: $12

Step-by-step explanation:

0.05 x 240 = $12

Hope this helps!

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Ldentify the following sections of the textbook.

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3 years ago

Cups are sold in packs and in boxes. There are 12 cups in each pack. There are 18 cups in each box.

valina [46]

Answer:

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3 years ago

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in a small school, there are 60 boys and 80 girls. write down the ratio of the number of boys to the number of girls. give your

tester [92]

60:80
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3 years ago

Point (-4,-8) Reflected across the x-axis

kkurt [141]

Answer:

(-4 , 8)

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The x-axis is the horizontal line. In this case, if you are reflecting over the x-axis, you are flipping the sign of the y-coordinate.

Same with the y-axis. If you are reflecting over the y-axis, you are flipping the sign of the x-coordinate.

In this case, reflect across the x-axis. Flip the sign of the y: (x , y)

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3 years ago

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

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

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3 years ago