Help pleaseee...A grocery store sells a bag of 7 oranges for $4.48. If Khloe spent $6.40 on oranges,

Mathematics

1 answer:

Kaylis [27]3 years ago

4 0

Answer:

Step-by-step explanation:

7 oranges over 4.48 dollars= x over 6.4 dollar

cross multiply and solve for x

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Jesse sold cupcakes and cookies yesterday. Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Je

ale4655 [162]

Answer: 19 cupcakes and 37 cookies were sold.

Step-by-step explanation:

Let x represent the number of cupcakes that Jesse sold yesterday.

Let y represent the number of cookies that Jesse sold yesterday.

If he sold 56 cupcakes and cookies combined, it means that

x + y = 56

Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Jesse had sold $37.75 worth of cookies and cupcakes. This means that

1.5x + 0.25y = 37.75 - - - - - - - - - - -1

Substituting x = 56 - y into equation 1, it becomes

1.5(56 - y) + 0.25y = 37.75

84 - 1.5y + 0.25y = 37.75

- 1.5y + 0.25y = 37.75 - 84

- 1.25y = - 46.25

y = - 46.25/ - 1.25

y = 37

x = 56 - y = 56 - 37

x = 19

7 0

3 years ago

Please explain this problem!!!

9966 [12]

tis a little of plain differentiation.

we know the radius of the cone is decreasing at 10 mtr/mins, or namely dr/dt = -10, decreasing, meaning is negative.

we know the volume is decreasing at a rate of 1346 mtr/mins or namely dV/dt = -1346, also negative.

so, when h = 9 and V = 307, what is dh/dt in essence.

we'll be needing the "r" value at that instant, so let's get it

$V=\cfrac{1}{3}\pi r^2 h\implies 307=\cfrac{\pi }{3}r^2(9)\implies \sqrt{\cfrac{307}{3\pi }}=r$

now let's get the derivative of the volume of the cone

$V=\cfrac{1}{3}\pi r^2 h\implies \cfrac{dV}{dt}=\cfrac{\pi }{3}\stackrel{product~rule}{ \left[ \underset{chain~rule}{2r\cdot \cfrac{dr}{dt}}\cdot h+r^2\cdot \cfrac{dh}{dt} \right]} \\\\\\ -1346=\cfrac{\pi }{3}\left[2\sqrt{\cfrac{307}{3\pi }}(-10)(9)~~+ ~~ \cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\right]$

$-\cfrac{4038}{\pi }=-\cfrac{180\sqrt{307}}{\sqrt{3\pi }}+\cfrac{307}{3\pi } \cdot \cfrac{dh}{dt}\implies \left[ -\cfrac{4038}{\pi }+\cfrac{180\sqrt{307}}{\sqrt{3\pi }} \right]\cfrac{3\pi }{307}=\cfrac{dh}{dt} \\\\\\ -\cfrac{12114}{307}+\cfrac{180\sqrt{3\pi }}{\sqrt{307}}=\cfrac{dh}{dt}\implies -7.920939735970634 \approx \cfrac{dh}{dt}$