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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"> <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]

f(x)=8x−6f(x)=8x-6 , [0,3][0,3]

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.(−∞,∞)(-∞,∞){x|x∈R}{x|x∈ℝ}f(x)f(x) is continuous on [0,3][0,3].f(x)f(x) is continuousThe average value of function ff over the interval [a,b][a,b] is defined as A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dx.A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dxSubstitute the actual values into the formula for the average value of a function.A(x)=13−0(∫308x−6dx)A(x)=13-0(∫038x-6dx)Since integration is linear, the integral of 8x−68x-6 with respect to xx is ∫308xdx+∫30−6dx∫038xdx+∫03-6dx.A(x)=13−0(∫308xdx+∫30−6dx)A(x)=13-0(∫038xdx+∫03-6dx)Since 88 is constant with respect to xx, the integral of 8x8x with respect to xx is 8∫30xdx8∫03xdx.A(x)=13−0(8∫30xdx+∫30−6dx)A(x)=13-0(8∫03xdx+∫03-6dx)By the Power Rule, the integral of xx with respect to xx is 12x212x2.A(x)=13−0(8(12x2]30)+∫30−6dx)

3 0

3 years ago