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3 years ago

Given the following sequences state the next number 1, 18, 16, 32 ____

Mathematics

1 answer:

8090 [49]3 years ago

8 0

Answer:

Not sure about the 1st sequence

Second sequence: 34

Third sequence: 18

Fourth sequence: 144

Step-by-step explanation:

The second sequence is the Fibonacci sequence. You find the next term by adding to two previous terms, so you have 13+21 which is equal to 34.

The third sequence is an arithmetic sequence that has a common difference of 11. Knowing this, we can add 11 to the last term to find the next one. This gives us 7+11 which is 18

The third sequence is the square of numbers starting form 16 and decreasing by one each time. This can be written with the formula $t_{n}$ = $(17-n)^2$ , where n is the term of the sequence you are solving for. Since you need to solve for the fifth term of the sequence the equation becomes, $t_{5}$ = $(17-5)^2$ which is the same as $13^2$ , so the 5th term is 169.

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Lin uses an app to graph the charge on her phone.

exis [7]

Answer:

See below

Step-by-step explanation:

<u>Use the graph to answer the following questions:</u>

When did she start using her phone?

At 2 afternoon

When did she start charging her phone?

At 8 afternoon

While she was using her phone, at what rate was Lin’s phone battery dying?

<u>From 100% to 40% between 2PM and 4 PM:</u>

(100 - 40)/(4 - 2) = 60/2 = 30% per hour

8 0

3 years ago

Read 2 more answers

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.