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professor190 [17]
3 years ago
5

https://lh3.googleusercontent.com/ZdagxxHvwERtpsxOHZwM5NSZgCwMCHN2KfRNp_BgcaHunkEAgxWxtlnBy9Kzw8sjh7CIuND8lTho9_dQLeH1AgXynwP1ro

UXC1RXLyfE-sKCs__gaY0wyzsKubzz=w307
Mathematics
2 answers:
DanielleElmas [232]3 years ago
8 0

What am I supposed to do?

lisov135 [29]3 years ago
5 0
It doesn’t let me click it
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If the measure of an angle is 38 degrees, find the measure of its complement.
Norma-Jean [14]
Answer: 52 degrees

Explanation:
For this problem, we must know that the complement of an angle is the measure of which will make a 90 degree sum.

So in this problem we simply solve the equation as follows:

38 + x = 90
x = 90 - 38
x = 52

Hence the complement of 38 is 52.

Cheers.
6 0
1 year ago
Spencer surveyed five of his friends to find out how many pets they have. His results are shown in the table below. What is the
Naily [24]

Answer:

6 pets

Step-by-step explanation:

5 0
2 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
2 years ago
Can someone tell me the formula i would use for this and how to solve through?
timama [110]
Since they are congruent, cpctc,
just set the congruent parts equal to each other. 
so 2x-20=30
and
15=3y-9
3 0
3 years ago
A=24 and c=26<br>what does b=____<br>show work please​
Oksi-84 [34.3K]

Answer:

f jb gfbjfmnb fjb

Step-by-step explanation:

f bmf bknmf

3 0
2 years ago
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