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marshall27 [118]
3 years ago
13

What is the perimeter?and how to I find it

Mathematics
2 answers:
Lelechka [254]3 years ago
5 0

Answer:

is there more info? I don't think this can be solved with given numbers could you take a more zoomed out photo?

Vladimir [108]3 years ago
4 0

Answer:

a permiter the continuous line forming the boundary of a closed geometric figure.

Step-by-step explanation:

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The map shows the location of the airport and a warehouse in a city. Though not displayed on the map, there is also a factory 64
Allushta [10]

warehouse location (24,-32)

airport location (-24,32)

 distance = sqrt((x2-x1)^2 +(y2-y1)^2)=

Sqrt(24--24)62 +(32--32)^2) =

sqrt(48^2 +64^2) =80

then distance from airport to the factory = -24 to 24 = 48

80+48 = 128 miles total

5 0
4 years ago
Read 2 more answers
Please answer all three.​
sasho [114]

Answer:

842.04 cm³

Step-by-step explanation:

Cylinder:

diameter = 6 cm

r = 6/2 = 3cm

h = 4 cm

Volume of cylinder = πr²h

                               = 3.14 * 3 * 3 * 4

                               = 113.04 cm³

Cube:

side = 9 cm

Volume of cube = side * side *side

                           = 9*9*9

                             = 729 cm³

Volume 0f composite figure = volume of cylinder + volume of cube

      = 113.04 + 729

      = 842.04 cm³

4 0
3 years ago
Read 2 more answers
Help me pls only 10points
NeX [460]

y=-7x+8

y=-7x-8

-7x+8=-7x-8

8=-8

x∈∅

B. No solutions

3 0
4 years ago
Need the answer for number 3 please!!!
Zepler [3.9K]

Answer:

Step-by-step explanation:

4 0
2 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
NNADVOKAT [17]

y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

4 0
3 years ago
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