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Dafna11 [192]
2 years ago
7

Helpppppppppppppppppppppppp

Mathematics
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:

a mean = 6 median = 7 mode = 8

b mean = 5 median = 5 mode= 5,7

c mean = 19 median = 18 mode= 22,18

d mean =5  median =5 mode= 5

Step-by-step explanation:

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Solve the following system of equations using the elimination method 5x - y = 10 x + 3y = 4​
Naddik [55]

Answer:

5x - y = 10x + 3y = 4

5x - y = 4..... 1

10x + 3y = 4.....2

multiply by -2 in eq 1 in both sides

-10x + 2y = -8

10x +3y = 4

eliminate one variable by adding the equations

5y = -4

y = - 4/5

sub the value of y in..1

5x - y = 4

5x + 4/5 = 4

5x = 16/5

divide by 5 on both sides

x = 16/25

(x ; y)

(16/25 ; -4/5)

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2 years ago
What is the value of x to the nearest tenth?
baherus [9]
Answer:
x
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Explanation:
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3 0
3 years ago
In each of Problems 7 through 11, determine whether the members of the given set of vectors are linearly independent. If they ar
Ad libitum [116K]

Answer:

Step-by-step explanation:

Check attachment

5 0
3 years ago
Which of the following are exterior angles? Check all that apply.
Marina86 [1]

Answer:

A

C

D

E

Step-by-step explanation:

Exterior angles can be described as the angles that are formed between the side of a polygon and the extended adjacent side of the polygon.

Or an exterior angle is the angle that is not inside the triangle formed.

The angles inside the triangle are interior angles.

Exterior angles are :

2

3

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Interior angles are :

1

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6 0
3 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
3 years ago
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