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romanna [79]
3 years ago
11

-2(4x-13)- -7=176 plzz hlp it was 10 points I'll make it 20

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0
The answer is -17.875. To solve it you have to make x by itself.
-2(4x-13)- -7=176
First, you would have to distribute.
-8x+26+7=176
Next, you combine like terms.
-8x+33=176
To get x by itself you next have to get rid of the constant.
-8x=143
Then you get x by itself by dividing the coefficient out.
x=-17.875
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After t seconds, a ball tossed in the air from the ground level reaches a height of h feet given by the equation h = 144t-16t^2
lesantik [10]
144t-16t^2= 16t(9-t)
after 3 seconds (just substitute t=3):
16(3)(9-3)=
16(3)(6)=
16*18=
288
maximum height: find the average of the roots:
the roots of 16t(9-t) are t=0 or t=9
since it's a parabola, the maximum is at t=4.5, at 324 ft
if its height is 224 feet, the equation is 224=144t-16t^2
16t^2-144t+224=0
divide by 16: t^2-9t+14=0
this can be factored as (t-2)(t-7)=0
the roots are t=2 and t=7, so the ball has been in the air for either 2 seconds or 7 seconds
the roots to 144t-16t^2 are 0 and 9, so the ball will hit the ground 9 seconds after being thrown
4 0
3 years ago
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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.

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