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4 years ago

4/7b=22 help solve please

1 answer:

7 0

I believe the answer is 77/2. I use this calculator to help me so it might be a little odd of the answer? Need an explanation?

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Suppose angle P is 110°. What would be the measure of arc LF? (Figure may not be drawn to scale)

lawyer [7]

If it is a central angle its 110 but if its an inscribed angle its 55

4 0

4 years ago

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igor_vitrenko [27]

Answer:

It's 105 square cm. I am sure

3 0

3 years ago

Read 2 more answers

Consider the diagram. Planes Q and R intersect. Lines d and c are vertical on plane R and form a right angle with line B on plan

harkovskaia [24]

Answer:

Line D or option C on edg

Step-by-step explanation:

Just took the pretest

7 0

3 years ago

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Which package has the lowest ratio of cost to number of lessons? $90 for 10 lessons $160 for 20 lessons $80 for 10 lessons $120

mario62 [17]

$90/10 = 9

$160/20 = 8

$80/10 = 10

$120/20 = 6

The package that has the lowest ratio cost to number of lessons is $120 for 20 lessons.

Hope this Helps!!

8 0

3 years ago

Read 2 more answers

Evaluate the given integral by changing to polar coordinates... \int \int_{D}^{}} xy dA ...where D is the disk with center the o

11111nata11111 [884]

Answer:

Step-by-step explanation:

If we use polar coordinates, the region D can be covered by replacing (x,y) by (r*sin(Θ),rcosΘ)), with 0<r<7, 0<Θ<2π. The differential matrix

$\left[\begin{array}{cc}rcos(\theta)&-rsin(\theta)\\sin(\theta)&cos(\theta)\end{array}\right]$

has determinant equal to r, so we can compute the double integral as follows

$\int\limits_D {xy} \, dx \, dy = \int\limits_0^{2\pi}\int\limits_0^r r^3cos(\theta)sin(\theta) \, dr \, d\theta$

(Note that we multiplied by the determinant of the Jacobian, r). A primitive for r³ is r⁴/4, thus, for Barrow's rule we have

$\int\limits_0^{2\pi}\int\limits_0^r r^3cos(\theta)sin(\theta) \, dr \, d\theta = \int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta$

A primitive of cos(Θ)sin(Θ) can be obtained using substitution, and it is sin²(Θ)/2 (note that the derivate of sin²(Θ) is 2sin(Θ)cos(Θ)). Therefore, taking both the dividing 4 and the 2 obtained, we have

$\int\limits_0^{2\pi}(\frac{r^4}{4}cos(\theta)sin(\theta)) \, |_{r = 0}^{r = 7} \, d\theta = \frac{1}{8} \int\limits_0^{2\pi} 7^4 * \frac{cos(\theta)sin(\theta)}{2} \, d\theta = \frac{7^4}{8} (sin^2(\theta)) |_{\theta=0}^{\theta=2\pi} \\= \frac{7^4}{8} (sin^2(2\pi)-sin^2(0)) = 0$

Hence, the integral is 0.

5 0

3 years ago