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Aleks04 [339]
3 years ago
6

Which expression is equivalent to 7 (2x - 5)

Mathematics
2 answers:
Anna [14]3 years ago
5 0

Answer:

B

Step-by-step explanation:

because it just is

zzz [600]3 years ago
3 0

Answer:

14x-35

Step-by-step explanation:

Use distributive property so,

you multiply 2x and -5 by 7

2x*7=14x

-5*7=-35

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Read 2 more answers
Factorise (2t-7)²-(5t-4)² <br>with working pls​
sertanlavr [38]

Answer:

33+12t−21t^2

Step-by-step explanation:

(2t-7)²-(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (2t-7)².

4t^2−28t+49−(5t-4)²

Use binomial theorem (a−b)^2 = a^2−2ab+b^2 to expand (5t-4)².

4t^2−28t+49−(25t^2−40t+16)

To find the opposite of 25t^2

−40t+16, find the opposite of each term.

4t^2−28t+49−25t^2−40t+16

Combine 4t^2  and −25t^2  to get −21t^2.

−21t^2−28t+49+40t−16

Combine −28t and 40t to get 12t.

−21t^2+12t+49−16

Subtract 16 from 49 to get 33.

−21t^2+12t+33

Swap terms to the left side.

33+12t−21t^2

I hope this helped!

5 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
4 years ago
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