Answer:
y=1/3, x=4/9
y=8, 32/3
x=9, y=27/4
y=20, x=80/3
Step-by-step explanation:
<span><span>f<span>(x)</span>=8x−6</span><span>f<span>(x)</span>=8x-6</span></span> , <span><span>[0,3]</span><span>[0,3]
</span></span>The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.<span><span>(−∞,∞)</span><span>(-∞,∞)</span></span><span><span>{x|x∈R}</span><span>{x|x∈ℝ}</span></span><span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuous on <span><span>[0,3]</span><span>[0,3]</span></span>.<span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuousThe average value of function <span>ff</span> over the interval <span><span>[a,b]</span><span>[a,b]</span></span> is defined as <span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>Substitute the actual values into the formula for the average value of a function.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8x−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8x-6dx)</span></span></span>Since integration is linear, the integral of <span><span>8x−6</span><span>8x-6</span></span> with respect to <span>xx</span> is <span><span><span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx</span><span><span>∫03</span>8xdx+<span>∫03</span>-6dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8xdx+<span>∫03</span>-6dx)</span></span></span>Since <span>88</span> is constant with respect to <span>xx</span>, the integral of <span><span>8x</span><span>8x</span></span> with respect to <span>xx</span> is <span><span>8<span>∫<span>30</span></span>xdx</span><span>8<span>∫03</span>xdx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>∫<span>30</span></span>xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(8<span>∫03</span>xdx+<span>∫03</span>-6dx)</span></span></span>By the Power Rule, the integral of <span>xx</span> with respect to <span>xx</span> is <span><span><span>12</span><span>x2</span></span><span><span>12</span><span>x2</span></span></span>.<span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>(<span><span>12</span><span>x2</span><span>]<span>30</span></span></span>)</span>+<span>∫<span>30</span></span>−6dx<span>)</span></span></span>
Answer:
Step-by-step explanation:
Let d represent the number of dimes. Then the number of quarters is 2d-3 and the total value of the coins is ...
0.10d + 0.25(2d-3) = 7.05
0.60d -0.75 = 7.05 . . . . . . . simplify
d = (7.05 +0.75)/0.60 = 13 . . . . add 0.75, divide by 0.60
2d-3 = 2·13 -3 = 23
Brandon has 23 quarters and 13 dimes.
