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tamaranim1 [39]
2 years ago
11

HELP ME PLEASEEE!!!!!!!

Mathematics
1 answer:
melisa1 [442]2 years ago
3 0

Answer:

The answer should be "Outlier".

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Determine the number of centimeters in 2 inches.
Paul [167]

Answer:

5

Step-by-step explanation:

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2 years ago
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Find the measure of x.
Andreyy89

Answer:

x = 10

Step-by-step explanation:

3 0
2 years ago
If you write out the numbers from 1 through 60 (including 1 and 60), how many numbers have a "5" in them?
kvasek [131]
Heya!!!
Answer is...

15 numbers have 5 in them....
Numbers are....
5
15
25
35
45
50
51
52
53
54
55
56
57
58
59


HOPE IT HELPS YOU '_'
7 0
3 years ago
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Please help me! I’m giving brainliest to whoever is right first
adelina 88 [10]

Answer:

y = 10

x = 17

Step-by-step explanation:

4y - 3 = 37

4y = 37 + 3

4y = 40

y = 40 / 4

y = 10

3x - 9 = 42

3x = 42 + 9

3x = 51

x = 51 / 3

x = 17

<em>Hope that helps!</em>

6 0
2 years ago
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Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-section
Angelina_Jolie [31]

Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

V = \int\limits^b_a {S(x)} \, dx          (1)

V = volume of the solid bounded by x = a and x = b

S(x) = cross section area of the solid, perpendicular to the x axis

From the figure we have that S is the area of a triangle that has base Z and height Y

Area of the triangle = S(x)=\frac{y(x)*z(x)}{2}          (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}          (3)

Equation of the line = y - y_{1} =m(x-x_{1} )          (4)

Replacing the points (1,0) and (0,1) in (3):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (4):

y-0=(-1)(x-1)

y(x) = -x + 1 (Line A-B)          (5)

We apply the equation of the point-slope line (plane xz):

Slope = m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}          (6)

Equation of the line = z - z_{1} =m(x-x_{1} )          (7)

Replacing the points (1,0) and (0,1) in (6):

m=\frac{1-0}{0-1} =-1

Replacing the point (1,0) and m = -1 in (7):

z-0=(-1)(x-1)

z(x) = -x + 1 (Line A-C)        (8)

Replacing (5) and (8) in (2)

S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}          (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx

V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)  evaluated from x=0 to x=1

V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}

3 0
3 years ago
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