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finlep [7]
3 years ago
8

Richard, Henry and Gavin share some sweets in the ratio 5:4:3. Richard gets 70 sweets. How many sweets are there altogether?

Mathematics
1 answer:
maxonik [38]3 years ago
4 0

There are 168 sweets altogether

<h3><u>Solution:</u></h3>

Richard, Henry and Gavin share some sweets in the ratio 5 : 4 : 3

Let "5a" be the share of Richard

Let "4a" be the share of Henry

Let "3a" be the share of Gavin

Richard gets 70 sweets

share of Richard = 70

5a = 70

<h3>a = 14</h3>

To find: number of sweets altogether

number of sweets altogether = share of richard + share of henry + share of gavin

number of sweets altogether = 5a + 4a + 3a

number of sweets altogether = 12a = 12(14) = 168

Thus there are 168 sweets altogether

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Answer:

x=-4\text{ and } x=3

Step-by-step explanation:

We are given the second derivative:

g''(x)=(x-3)^2(x+4)(x-6)

And we want to find its inflection points.

To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.

Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.

So first, finding the zeros, we see that:

0=(x-3)^2(x+4)(x-6)\Rightarrow x=-4, 3, 6

So, we can draw the following number-line:

<----(-4)--------------(3)----(6)---->

Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.

Testing for x < -4, we can use -5. So:

g^\prime^\prime(-5)=(-5-3)^2(-5+4)(-5-6)=704>0

Since we acquired a positive result, g(x) is concave up for x < -4.

For -4 < x < 3, we can use 0. So:

g^\prime^\prime(0)=(0-3)^2(0+4)(0-6)=-216

Since we acquired a negative result, g(x) is concave down for -4 < x < 3.

And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.

For 3 < x < 6, we can use 4. So:

g^\prime^\prime(4)=(4-3)^2(4+4)(4-6)=-16

Since we acquired a negative result, g(x) is concave down for 3 < x < 6.

Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.

And finally, for x > 6, we can use 7. So:

g^\prime^\prime(7)=(7-3)^2(7+4)(7-6)=176>0

So, g(x) is concave up for x > 6.

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Back: 15.6 x 13.3 = 207.48

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