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makkiz [27]
3 years ago
9

A rectangles width is 5 feet less than its length. Write a quadratic function that expresses the rectangles area in terms of its

length
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
4 0

Answer:

Area = L²-5L

Step-by-step explanation:

Area = length * width

Length = L

Width = L - 5

Area = L * (L-5) = L²-5L

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3(2x + 11) and (3x + 15)(2)
algol [13]

Answer:

3(2x+11)= 6x+33  and (3x+15)(2) = 6x+30

Step-by-step explanation:

Use distributive property to multiply the outside factor to each factor inside the parenthesis.  

3(2x+11)

(3*2x)+(3*11)

6x+33

3 0
3 years ago
A 445-foot fence encloses a pasture. What is the length of each side of the pasture?
fredd [130]

Answer:

89*5

Step-by-step explanation:

3 0
3 years ago
If Subway sells turkey sandwiches for $3.79 each, how many sandwiches can you buy for your 6 friends to share with $20.00
Sophie [7]
3.75 each, 6 friends, $20. 6 multiplied by 3.75 is 18.75. you can buy 5 sandwiches
4 0
3 years ago
Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an
Natasha_Volkova [10]

Answer:

The volume of the solid is 243\sqrt{2} \ \pi

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0  ≤ ρ²   ≤   81

0  ≤ ρ   ≤  9

The intersection that takes place in the sphere and the cone is:

x^2 +y^2 ( \sqrt{x^2 +y^2 })^2  = 81

2(x^2 + y^2) =81

x^2 +y^2 = \dfrac{81}{2}

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

z = \sqrt{x^2+y^2}

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; \dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}

Thus, volume: V  = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4}  \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho   ^2 \ sin \phi \ d\rho \   d \theta \  d \phi

V  = \int \limits^{\pi/2}_{\pi/4} \ sin \phi  \ d \phi  \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho   ^2 d\rho

V = \bigg [-cos \phi  \bigg]^{\pi/2}_{\pi/4}  \bigg [\theta  \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3}  \bigg ]^{9}_{0}

V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]

V = 243\sqrt{2} \ \pi

4 0
3 years ago
Identify the initial value and rate of change for the graph shown. (4 points)
spin [16.1K]

Answer:

a

Step-by-step explanation:

8 0
3 years ago
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