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3 years ago

Solve the question and check the solution. Express numbers as integers or simplified fractions. 5(m+5)=7m+9

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

Answer:

m = 8

Step-by-step explanation:

5(m+5)=7m+9 (distribute & evaluate parenthesis)

5(m)+5(5)=7m+9

5m + 25 = 7m + 9 (move all m-terms to one side and constants to the other)

5m - 7m = 9 - 25 (evaluate both sides)

-2m = -16 (divide both sides by -2)

m = (-16) / ( -2) = 8

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If AB = x+ 4, BC = 2x -10 and AC

lawyer [7]

Answer:

Step-by-step explanation:

By the diagram, you can see the sum of segment AB and segment BC is segment AC. Adding the given expressions for AB and B is $3x-6$ . Simplifying the equation $3x-6=2x+1$ , gives $x=7$ . Substituting $x=7$ in the equation for segment AC gives 15.

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2 years ago

Which of the following pairs of points are both solutions to the equation 2x-5y=6 ? and and and and

zhannawk [14.2K]

Answer:

Any points on the line y=2/5 x - 6/5. See picture below.

Step-by-step explanation:

Convert the equation to slope intercept form and graph it.

2x - 5y = 6

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y = -2/-5 x + 6/-5

y=2/5 x - 6/5.

Locate each of the points listed on the graph. If they are a part of the line, then they are solutions.

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3 years ago

The hypotenuse of the right triangle is 1 in. more than twice the longer leg. The length of the shorter leg is √10 in. Find the

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2 years ago

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:

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

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3 years ago

Powers of 10 with positive exponents are: less than 1 greater than 1

nata0808 [166]

Answer:

greater than 1

Step-by-step explanation:

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3 years ago

Read 2 more answers