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Musya8 [376]
3 years ago
13

Solve for m.

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

a

Step-by-step explanation:

solve for m. ;)

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Write a number that has a 2 with a value 100 times less than the value of the 2 in the number 42,845
bagirrra123 [75]
The answer is 20. The value of the digit 2 in 42,845 is 2,000. 100 times less than 2,000 is 20. 20 has a 2 in it.
6 0
2 years ago
Otto used 6 cups of whole wheat flour and x cups of white flour in the recipe. What is the equation that can be used to find the
Nezavi [6.7K]

Answer:

Well, there are x amounts of white flower and 6 cups of wheat flower.

So the total flower is x + 6

Given that  is the total, the equation you would use is:

y=x+6

The constraints are as follows:

y can only be > 6

And if y=0, x would have to be -6 (which is impossible) 

3 0
2 years ago
What is the missing side equal to?
Rom4ik [11]
This is a right triangle and you can use the Pythagorean theorem to find the missing side. 

5^2 + 12^2 = c^2
25 + 144 = c^2
169 = c^2
square root of 169 = c
13.

So the missing side is 13m
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
2 years ago
What is the solution to this equation x+12= -5
scZoUnD [109]
-17, I think.
12 - -5 = -17
17/1 =-17
X=-17
5 0
3 years ago
Read 2 more answers
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