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3 years ago

Chapter 1: Quadrati

Mathematics

1 answer:

Bad White [126]3 years ago

8 0

Answer:

mid point of AB (0.5 , 0)

Step-by-step explanation:

2x + 5y = 1 2x = 1 - 5y

y = 2xy+ 5 = (1 - 5y)*y + 5 = y - 5y² + 5

y-y = y - 5y² + 5 - y = - 5y² + 5

0 = - 5y² + 5

5y² = 5

y² = 1

y = 1 or y = -1

if y = 1 x = 1/2 (1 - 5) = -2 .... point A (-2 , 1)

if y = -1 x = 1/2 (1 + 5) = 3 ..... point B (3 , -1)

mid point of AB (x' , y') : x' = 1/2 (x₁ + x₂) = 1/2 (-2 + 3) = 1/2

y' = 1/2 (y₁ + y₂) = 1/2 (-1 + 1) = 0

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3 years ago

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Hello!

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❖ To find the mean, add up all the numbers and divide by the amount of numbers there are. To find the range, subtract the highest number by the lowest number in the data set.

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~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

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3 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

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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2 years ago