X/7 ≥ -6 solve for x

aniked [119]

-40

Explanation:
(-7+2) would equal -5, then you multiply -5 by 8 and you get -40.

3 0

3 years ago

Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l

SIZIF [17.4K]

Answer:

$\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}$

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let

$C_1,C_2$

be the two paths.

Recall that if we parametrize a path C as $(r_1(t),r_2(t),r_3(t))$ with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt$

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and

$\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}$

The parametrization of the line segment from (2,2,2) to

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)

r'(t) = (-11,4,3)

and

$\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}$

Hence

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}$

8 0

3 years ago

graph the function f (x) = x^2 + 4x -5. on the coordinate plane. (a) What are the x-intercepts? (b) What are the y-intercepts? (

Butoxors [25]

Answer:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

See attached graph.

Step-by-step explanation:

To graph the function, find the vertex of the function find (-b/2a, f(-b/2a)). Substitute b = 4 and a = 1.

-4/2(1) = -4/2 = -2

f(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -4 - 5 = -9

Plot the point (-2,-9). Then two points two points on either side like x = -1 and x = -3. Substitute x = -1 and x = -3

f(-1) = (-1)^2 + 4 (-1) - 5 = 1 - 4 - 5 = -8

Plot the point (-1,-8).

f(-3) = (-3)^2 + 4(-3) - 5 = 9 - 12 - 5 = -8

Plot the point (-3,-8).

See the attached graph.

The features of the graph are:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

5 0

3 years ago

Angelica had 8 more pairs of shoes than her mom. Together they have 30 pairs of shoes. How many pair of shoes does each one have

MAVERICK [17]

Answer:

The answer to your question is Angelica 19, her mom = 11

Step-by-step explanation:

Data

Angelica has 8 more than her mother

total number of shoes = 30 pairs

The Number of pairs Angelica has = a = ?

The Number of pairs her mother has = m = ?

Process

1.- Write two equation that help to solve this problem

a = m + 8 Equation l

a + m = 30 Equation ll

2.- Solve the system of equations by substitution

Substitute equation l in equation ll

(m + 8) + m = 30

-Solve for m

m + 8 + m = 30

2m = 30 - 8

2m = 22

m = 22/2

m = 11

3.- Find a

a = 11 + 8

a = 19

4.- Conclusion

Angelica has 19 pairs of shoes and her mom has 11

5 0

3 years ago

A building makes a 90° angle with the ground. A ladder leans against the building, making a 110° exterior angle with the ground.

liberstina [14]

<u>Answer:</u>

A building makes a $90^{\circ}$ angle with the ground .The two interior angles are $70^{\circ}$ and $20^{\circ}$

<u>Solution: </u>

Given, a ladder which is against the building wall makes an exterior angle of $110^{\circ}$

And building makes a $90^{\circ}$ angle with the ground.

So, altogether it makes a right angle triangle with ladder as hypotenuse and ground as base leg and building as another leg.

Now, we know that, sum of exterior angles and interior angles equals to $180^{\circ}$

Here, in the case of ground and ladder, exterior angle is $110^{\circ}$ and interior angle is unknown.

Exterior angle + interior angle = 180

110 + interior angle = 180

Interior angle = 180 – 110

Interior angle = $70^{\circ}$

We have found one of the two interior angles of right angle triangle.

We know that, sum of angles in a triangle is 180 degree

Known Interior angle + unknown interior angle + right angle = 180

70 + unknown interior angle + 90 = 180

Unknown interior angle + 160 = 180

Unknown interior angle = 180 - 160

Unknown interior angle = $20^{\circ}$

Hence the two interior angles are $70^{\circ}$ and $20^{\circ}$

7 0

4 years ago