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3 years ago

6

According to the presentation, what are situations that can cut into retirement funds? accidents tax liabilities health needs So

cial Security not being properly funded

1 answer:

ivanzaharov [21]3 years ago

8 0

Accidents, health needs.

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Um how can i do exponential decay

laila [671]

What do you mean ???

4 0

3 years ago

Read 2 more answers

Gabe Amodeo, a nuclear physicist, needs 60 liters of a 60% acid solution. He currently has a 50% solution and a 70% solution.

In-s [12.5K]

Answer:

Gabe needs 20 liters of the 40% solution.

He also needs 40 liters of the 70% solution.

Step-by-step explanation:

7 0

3 years ago

Provide an example of a trig function that includes multiple transformations. Describe how it is different from the standard tri

klio [65]

An example of a trig function that includes multiple transformations and how it is different from the standard trig function is; As detailed below

<h3>How to interpret trigonometric functions in transformations?</h3>

An example of a trigonometric function that includes multiple transformations is; f(x) = 3tan(x - 4) + 3

This is different from the standard function, f(x) = tan x because it has a vertical stretch of 3 units and a horizontal translation to the right by 4 units, and a vertical translation upwards by 3.

Another way to look at it is by;

Let us use the function f(x) = sin x.

Thus, the new function would be written as;

g(x) = sin (x - π/2), and this gives us;

g(x) = sin x cos π/2 - (cos x sin π/2) = -cos x

This will make a graph by shifting the graph of sin x π/2 units to the right side.

Now, shifting the graph of sin xπ/2 units to the left gives;

h(x) = sin (x + π/2/2)

Read more about Trigonometric Functions at; brainly.com/question/4437914

#SPJ1

5 0

2 years ago

Use induction to show that 12 + 22 + 32 + ... + n2 = n(n+1)(2n+1)/6, for all n > 1.

dlinn [17]

Answer with Step-by-step explanation:

Let $P(n)=1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

Substitute n=2

Then $P(2)=1+2^2=5$

$P(2)=\frac{2(2+1)(4+1)}{6}=5$

Hence, P(n) is true for n=2

Suppose that P(n) is true for n=k >1

$P(k)=1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, we shall prove that p(n) is true for n=k+1

$P(k+1)=1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS

$P(k+1)=1^2+2^2+3^2+.....+k^2+(k+1)^2$

Substitute the value of P(k)

$P(k+)=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

$P(k+1)=(k+1)(\frac{k(2k+1}{6})+k+1)$

$P(k+1)=(k+1)(\frac{2k^2+k+6k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+7k+6}{6})$

$P(k+1)=(k+1)(\frac{2k^2+4k+3k+6}{6})$

$P(k+1)=\frac{(k+1)(k+2)(2k+3)}{6}$

LHS=RHS

Hence, P(n) is true for all n >1.

Hence, proved

4 0

3 years ago

I have no idea what im doing, can someone please help me? ASAP!!!! Use the pythagorean theorem to find the missing slide.

Solnce55 [7]

Answer:

1. b=√56

Step-by-step explanation:

1. Pythagorean therom states a²+b²=c² and in number 1 we are given the c or the hypotenuse so plugging in the values we know 13²+b²=15²

which equals 169+b²=225 subtract 169 from both sides and we get b²=56

square root both sides and we get b=√56

7 0

3 years ago