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shepuryov [24]
3 years ago
11

Mariah spent $9.50 on 9 pounds of limes and pears. Limes cost $0.50 per pound and pears cost $1.50 per pound. Let l be the numbe

r of pounds of limes and let p be the number of pounds of pears. The system of linear equations that models this scenario is: l + p = 9 0.5l + 1.5p = 9.5 How many pounds of each type of fruit did she buy?
She bought ______ pounds of limes and ______ pounds of pears.
Mathematics
1 answer:
Norma-Jean [14]3 years ago
7 0
She bought 4 pounds of limes and 5 pounds of pears.
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Can I get help please and thank you
Ipatiy [6.2K]

9514 1404 393

Answer:

  a.  0.81

  b.  v = 28000(0.81^n)

  c.  2757.36

Step-by-step explanation:

a. The growth factor is 1 more than the growth rate. Here, the growth rate is -19% (per year), so the growth factor, the multiplier, is ...

  1 -0.19 = 0.81

__

b. The equation sets value equal to the original value multiplied by the growth factor to the power of the number of years:

  value = (original value) × (growth factor)^n

  v = 28000(0.81^n)

__

c. For n=11, this is ...

  v = 28000(0.81^11) ≈ 2757.36

The value of the truck after 11 years is about $2757.

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3 years ago
I need help ASAP there’s a time limit
andreyandreev [35.5K]
14 uehebeudygwgwvbsbbe
3 0
3 years ago
Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the
dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

I = \int \int_E \int zdV where E is bounded by the cylinder y^2 + z^2 = 81 and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and y^2 + z^2 = 81

∴ z^2 = 81 - y^2

z = \sqrt{81 - y^2}

Thus, z lies between 0 to \sqrt{81 - y^2}

GIven curve x = 0 and y = 9x

x =\dfrac{y}{9}

As such,x lies between 0 to \dfrac{y}{9}

Given curve x = 0 , x =\dfrac{y}{9} and z = 0, y^2 + z^2 = 81

y = 0 and

y^2 = 81 \\ \\ y = \sqrt{81}  \\ \\  y = 9

∴ y lies between 0 and 9

Then I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix}    ^ {\sqrt {{81-y^2}}}_{0} \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{(\sqrt{81 -y^2})^2 }{2}-0  \end {bmatrix}     \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{{81 -y^2} }{2} \end {bmatrix}     \ dxdy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0}    \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix}     \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81 \  y -y^3} }{18} \end {bmatrix}     \ dy

I = \dfrac{1}{18} \int^9_{y=0}  \begin {bmatrix}  {81 \  y -y^3}  \end {bmatrix}     \ dy

I = \dfrac{1}{18}  \begin {bmatrix}  {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}}  \end {bmatrix}^9_0

I = \dfrac{1}{18}  \begin {bmatrix}  {40.5 \ (9^2) - \dfrac{9^4}{4}}  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  3280.5 - 1640.25  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  1640.25  \end {bmatrix}

I = 91.125

4 0
3 years ago
For Madeline's lemonade recipe, 3 lemons are required to make 6 cups
expeople1 [14]

Answer:

1/2 lemon per 1 cup

Step-by-step explanation:

6/6=1

3/6=1/2= .5 lemon

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3 years ago
Which set of points represents a function?
Vesnalui [34]
Is there a picture for this question or options ?
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3 years ago
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