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shepuryov [24]
3 years ago
10

What is the perimeter of a rectangle with a length of 16cm and a width of 12cm?

Mathematics
2 answers:
Bumek [7]3 years ago
8 0

Answer:

56

Step-by-step explanation:

Add 16 and 16 then add 12 and 12 and the results together and there is your answer

Tanya [424]3 years ago
3 0

Answer:

4cm is the rectangle

Step-by-step explanation:

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Which equation matches the graph below?
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Answer:Given that the graph shows tha the functión at x = 0 is below the y-axis, the constant term of the function has to be negative. This leaves us two possibilities:

y = 8x^2 + 2x - 5 and y = 2x^2 + 8x - 5

To try to discard one of them, let us use the vertex, which is at x = -2.

With y = 8x^2 + 2x - 5, you get y = 8(-2)^2 + 2(-2) - 5 = 32 - 4 - 5 = 23 , which is not the y-coordinate of the vertex of the curve of the graph.

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Step-by-step explanation:

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0
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So, the definite integral  \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Given that

\int\limits^1_0 {x^{2} } \, dx = 13

We find

\int\limits^1_0 {(4 - 6x^{2} )} \, dx

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4    - 6\int\limits^1_0 {x^{2} } \, dx

Since

\int\limits^1_0 {x^{2} } \, dx = 13,

Substituting this into the equation the equation, we have

\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Learn more about definite integrals here:

brainly.com/question/17074932

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Step-by-step explanation:

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