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Len [333]
3 years ago
9

SMARTIES PLEASE HELP

Mathematics
2 answers:
ELEN [110]3 years ago
7 0
1= 5
2=15
3=25
I think I’m not sure tho
bogdanovich [222]3 years ago
4 0
1. 5
2. 15
3. 25
i believe
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Find the values of x and y.
kotykmax [81]

Answer:

y=9 and x=18

Step-by-step explanation:

You know that y+3=12 so 12-3=9

You know that x-3=15 so 15+3=18

6 0
3 years ago
Can someone plz help me with this one problem plzzzzz!!!!
Alinara [238K]
Again lol yes the answer is yea
7 0
3 years ago
Read 2 more answers
If u know this help me I got a lot of homework 6 grade work
Svet_ta [14]
Hello! I can help you!

22. Okay. What we need to do for this one is subtract fraction.l, because we are looking for how much more the second container holds than the first. We need to covert them to like denominators, which in this case is 8 5 3/4 = 5 6/8. Because 6/8 is less than 7/8, we need to regroup by adding 8 and turning the 5 into a 4. The problem would become like this: 4 14/8 - 1 7/8. When you subtract both numbers, you get 3 7/8. There. The second container holds 3 7/8 more gallons of water than the first.

23.

a. 15 student speeches will last 1 1/2 minutes each. In this case, we just simply multiply the fractions by multiply straight across. Multiply the top numbers together and the bottom numbers together. 1 1/2 is 3/2 a an improper fraction. Make 15 have a denominator of 1. 15/1 * 3/2 is 45/2 or 22 1/2 as a mixed number. It will take 22 1/2 minutes to give the speeches.

b. Okay. The teacher makes a 15 minute intro. Let’s add the time it takes to make the speeches and the teacher’s intro. 22 1/2 + 15 0/2 is 37 1/2. Yes. There is enough time to record everyone.

c. There are 60 minutes in 1 hour. Subtract the total amount taken on the camera from 60. 0/2 is less than 1/2. Again, regroup by adding 2 to make 2/2, and crossing out 60, so it becomes 59. 59 2/2 - 37 1/2 is 22 1/2. There are 22 1/2 minutes left on the digital camera.
5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20" id="TexFormula1" title=" " alt=" " align="absmiddle" class="latex-formula"><br><img src="
seropon [69]
\text{The range of}\ y=-3^x\ \text{is}\ (-\infty;\ 0).\\\\\text{The range of}\ f(x)=-3^x+7\ \text{is}\ (-\infty;\ 7)
6 0
3 years ago
How to find average value of a function over a given interval?
Strike441 [17]
<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>
3 0
3 years ago
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