Answer:
(x, y) = (-4, 15)
Step-by-step explanation:
The two equations have the same coefficient for y, so you can eliminate y by subtracting one equation from the other. Here the x coefficient is largest for the first equation, so it will work best to subtract the second equation.
(3x +y) -(2x +y) = (3) -(7)
x = -4 . . . . . . . . simplify
Now, we can find y by substituting this value for x.
2(-4) +y = 7
y = 7 +8 = 15 . . . . . add 8 to both sides of the equation
The solution is (x, y) = (-4, 15).
This is an EXPONENT.
To do this all we need to do is multiply the base by itself how many times the power says.
(-6)² = (-6) × (-6) = 36
Recall that any negative number times another negative number is positive so that is why this would be positive.
So, (-6)² = 36
Hope I helped ya!!
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
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For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
Pemdas 1. parentheses: everything is already in parentheses. so next is exponents. 10×-2 is -20. and 10×6 is 60. so 4×-20 is the next step. it equals -80. and then 3×60 is 180. lastly,add -80 and 180 together. the answer should be 100 if i did that correctly.
115 = 10X + 0.25(60)
115 = 10x +15 (this is the equation to use)
100=10x
x=10 hours
