Answer:
Yes.
2 feet
Step-by-step explanation:
Pieces of copper that Arsenio needs in total = 27 pieces
Length of each piece of copper tubing that he needs = 2/9 foot
Length of available tubing = 8 foot
Since, the length of each piece is 2/9 foot, the length of 27 pieces would be:
This means, Arsenio needs 6 feet in total for the copper tubing. Since he got 8 feet in total, he has sufficient amount of tubing available.
After making the 27 pieces, he will be left with 8 - 6 = 2 feet of tubing.
Answer:
Step-by-step explanation:
<u>Use formula:</u>
- A = 1/2 Pa, where P-perimeter of the polygon, a- apothem
1)
- A = 1/2(7*13.9)(14.4) = 700.56 ≈ 700.6
Correct choice is B
2)
Correct choice is A
Answer:
El chico nació en el año 18 antes de cristo.
La madre nació en el año 40.
Step-by-step explanation:
Las respuestas son fáciles:
- Si en el año 5, el hijo tiene 13 años, hagamos que a partir del año 0 después de cristo sea positivo.
-13-(-5) = -18
- Madre:
-18-22 = -40
Entonces, su edad debería ser negativa, simplemente cambiamos el símbolo y lo pasamos a positivo.
The area of the trampoline is mathematically given as
a = 3.14
<h3>What is the area?</h3>
Question Parameters:
You found one online that has a circumference of 6.28 yards
Generally, the equation for the Circumference is mathematically given as
C = 2*pi*r
Thererfore
6.28 = 2*3.14*r
r = (6.28)/(6.28)
r = 1
In conclusion, where radius is
a = 3.14*1^2
a = 3.14
Read more about Arithmetic
brainly.com/question/22568180
<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>
