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3 years ago

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Can you help me with my maths?

1 answer:

Assoli18 [71]3 years ago

3 0

Answer:

45. B, 9 + 0 = 9

46. D. 9 x 1 = 9

47. D. 0

48. B. 1/6 x 6 = 1

49. A

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

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

4 0

2 years ago

PLEASE HELP!!!!!

Ne4ueva [31]

Answer:

Equation: $30=2(x-2)+2(2x+3)$

$x:\frac{14}{3}$

Step-by-step explanation:

By definition, the perimeter of a figure can be calculated by adding the lenghts of its sides.

Knowing this, you can write the following equation:

$P=2(x-2)+2(2x+3)$ <em> [Equation 1]</em>

According to the data given in the exercise, the perimeter in feet of the fighter is:

$P=30$

Therefore, you can substitute this values into<em> [Equation 1]:</em>

$30=2(x-2)+2(2x+3)$

Finally, you must solve for "x" in order to find its value. This is:

$30=6x+2\\\\30-2=6x\\\\\frac{28}{6}=x\\\\x=\frac{14}{3}$

3 0

2 years ago

The length of a rectangle is 4 less than twice the width. If the perimeter of the rectangle is 130, find the length of the recta

Alchen [17]

Answer:

Let L be the length of the rectangle and w be the width of the rectangle.

"The lenght of a rectangle is one less than twice the width" means L=2W-1

Using perimeter formula of a rectangle which is P=2(L+W) you have:

P=2(L+W)

130=2(2W-1+W)

Solve equation above to find W

130=2(3W-1)

130=6W-2

130+2=6W

132=6W

W=22

From here you find L=2W-1=2x22-1=43

Only thing left is to find area A=LxW

4 0

3 years ago

X/16-1/4=1/2 solve for x

Alina [70]

x=12

x/16-1-4=1/2

multiply by both sides

x-4=8

move constant to the right side of and change its sign

x=8+4

add the numbers

x=12

4 0

3 years ago

13 in<br> 10 in.<br> 7 in<br> 3. What is the perimeter of the<br> (19)<br> trapezoid?

quester [9]

37 because you add 13+10=23 7+7=14 and 23+14=37

5 0

2 years ago