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2 years ago

Please help me and explain the anwser

Mathematics

1 answer:

natita [175]2 years ago

8 0

ANSWER: y=7 is the equation Of the line that passes to through the point (-2,7) and Has a slope of zero.

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In the following calculation, which part should be calculated first? 3 + (2 * 4) - 5 / 2

meriva

Always solve the parenthese first

8 0

3 years ago

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Which graph represents a function with direct variation?

sweet [91]

This is easier than you think!

When a graph shows direct variation, it always passes through the origin

The origin is (0, 0)

The very center!

Your answer is the first graph!

5 0

3 years ago

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37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

3 years ago

For which value of x does f(x) = 2

laila [671]

Answer:

Step-by-step explanation:

Look a the picture.

Draw the horizontal line y = 2.

Common point: (-4, 2).

Conclusion: f(x) = 2 → x = -4

8 0

3 years ago

An image of a parabolic lens is projected onto a graph. The y-intercept of the graph is (0, 90), and the zeros are 5 and 9. Whic

seraphim [82]

Given:

y-intercept of the graph: (0, 90)
zeros: 5 and 9

The equation that models the function based on the zeros given, is either

y = 90 (x-5) (x-9)
or
y= 2(x-5)(x-9)

try solving for the y-intercept of each function,

y = 90 (0-5) (0-9)
y = 4050
(0, 4050)

y = 2(0-5) (0-9)
y = 90
(0, 90)

therefore, the equation that models the function is y = 2(x-5)(x-9)

8 0

3 years ago

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