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2 years ago

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Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the neares

t hundredth.
1.)a17 ≈ 123,802.31
2.)a17 ≈ 30,707.05
3.a17 ≈ 19,684.01
4.)a17 ≈ 216,654.05

2 answers:

Mashutka [201]2 years ago

8 0

The geometric sequence formula is expressed as an = a1 * r^(n-1) where n is an integer. In this case, upon substitution, 150.06 = 16 * r^(4). extracting r, r is equal to 1.75. Hence the 17th term from the formula is equal to 123802.32.

Lunna [17]2 years ago

6 0

Answer:

The correct option is 1.

Step-by-step explanation:

It is given that the first term of a geometric sequence is 16 the fifth term of the sequence is 150.06.

$a_1=16$

$a_5=150.06$

The nth term of a geometric sequence is

$a_n=a_1r^{n-1}$ .... (1)

The fifth term of the sequence is

$a_5=a_1r^{5-1}$

Substitute $a_1=16$ and $a_5=150.06$ .

$150.06=16r^{4}$

Divide both sides by 16.

$9.37875=r^{4}$

$(9.37875)^{\frac{1}{4}}=r$

$r\approx 1.75$

Substitute n=17, $a_1=16$ and $r=1.75$ to find the 17th term.

$a_{17}=16(1.75)^{17-1}$

$a_{17}=123802.31384$

$a_{17}\approx 123802.31$

Therefore the correct option is 1.

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12 catsate 40 pounds of cat food.At this rate, how many pounds of cat food would 48 cats eat?

max2010maxim [7]

The answer is 160, multiply 12 by 4 to get 48 cats. 40 multiplied by 4 = 160.
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2 years ago

The center of a circle is at the origin on a coordinate grid. The vertex of a parabola that opens upward is at (0, 9). If the ci

zhannawk [14.2K]

Answer:

"The maximum number of solutions is one."

Step-by-step explanation:

Hopefully the drawing helps visualize the problem.

The circle has a radius of 9 because the vertex is 9 units above the center of the circle.

The circle the parabola intersect only once and cannot intercept more than once.

The solution is "The maximum number of solutions is one."

Let's see if we can find an algebraic way:

The equation for the circle given as we know from the problem without further analysis is so far $x^2+y^2=r^2$ .

The equation for the parabola without further analysis is $y=ax^2+9$ .

We are going to plug $ax^2+9$ into $x^2+y^2=r^2$ for $y$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

To expand $(ax^2+9)^2$ , I'm going to use the following formula:

$(u+v)^2=u^2+2uv+v^2$ .

$(ax^2+9)^2=a^2x^4+18ax^2+81$ .

$x^2+y^2=r^2$

$x^2+(ax^2+9)^2=r^2$

$x^2+a^2x^4+18ax^2+81=r^2$

So this is a quadratic in terms of $x^2$

Let's put everything to one side.

Subtract $r^2$ on both sides.

$x^2+a^2x^4+18ax^2+81-r^2=0$

Reorder in standard form in terms of x:

$a^2x^4+(18a+1)x^2+(81-r^2)=0$

The discriminant of the left hand side will tell us how many solutions we will have to the equation in terms of $x^2$ .

The discriminant is $B^2-4AC$ .

If you compare our equation to $Au^2+Bu+C$ , you should determine $A=a^2$

$B=(18a+1)$

$C=(81-r^2)$

The discriminant is

$B^2-4AC$

$(18a+1)^2-4(a^2)(81-r^2)$

Multiply the (18a+1)^2 out using the formula I mentioned earlier which was:

$(u+v)^2=u^2+2uv+v^2$

$(324a^2+36a+1)-4a^2(81-r^2)$

Distribute the 4a^2 to the terms in the ( ) next to it:

$324a^2+36a+1-324a^2+4a^2r^2$

$36a+1+4a^2r^2$

We know that $a>0$ because the parabola is open up.

We know that $r>0$ because in order it to be a circle a radius has to exist.

So our discriminat is positive which means we have two solutions for $x^2$ .

But how many do we have for just $x$ .

We have to go further to see.

So the quadratic formula is:

$\frac{-B \pm \sqrt{B^2-4AC}}{2A}$

We already have $B^2-4AC}$

$\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}$

This is t he solution for $x^2$ .

To find $x$ we must square root both sides.

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

So there is only that one real solution (it actually includes 2 because of the plus or minus outside) here for x since the other one is square root of a negative number.

That is,

$x=\pm \sqrt{\frac{-(18a+1) \pm \sqrt{36a+1+4a^2r^2}}{2a^2}}$

means you have:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

or

$x=\pm \sqrt{\frac{-(18a+1)-\sqrt{36a+1+4a^2r^2}}{2a^2}}$ .

The second one is definitely includes a negative result in the square root.

18a+1 is positive since a is positive so -(18a+1) is negative

2a^2 is positive (a is not 0).

So you have (negative number-positive number)/positive which is a negative since the top is negative and you are dividing by a positive.

We have confirmed are max of one solution algebraically. (It is definitely not 3 solutions.)

If r=9, then there is one solution.

If r>9, then there is two solutions as this shows:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}}$

r=9 since our circle intersects the parabola at (0,9).

Also if (0,9) is intersection, then

$0^2+9^2=r^2$ which implies r=9.

Plugging in 9 for r we get:

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+4a^2(9)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{36a+1+324a^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+\sqrt{(18a+1)^2}}{2a^2}}$

$x=\pm \sqrt{\frac{-(18a+1)+18a+1}{2a^2}}$

$x=\pm \sqrt{\frac{0}{2a^2}}$

$x=\pm 0$

$x=0$

The equations intersect at x=0. Plugging into $y=ax^2+9$ we do get $y=a(0)^2+9=9$ .

After this confirmation it would be interesting to see what happens with assume algebraically the solution should be (0,9).

This means we should have got x=0.

$0=\frac{-(18a+1)+\sqrt{36a+1+4a^2r^2}}{2a^2}$

A fraction is only 0 when it's top is 0.

$0=-(18a+1)+\sqrt{36a+1+4a^2r^2}$

Add 18a+1 on both sides:

$18a+1=\sqrt{36a+1+4a^2r^2$

Square both sides:

$324a^2+36a+1=36a+1+4a^2r^2$

Subtract 36a and 1 on both sides:

$324a^2=4a^2r^2$

Divide both sides by $4a^2$ :

$81=r^2$

Square root both sides:

$9=r$

The radius is 9 as we stated earlier.

Let's go through the radius choices.

If the radius of the circle with center (0,0) is less than 9 then the circle wouldn't intersect the parabola. So It definitely couldn't be the last two choices.

7 0

2 years ago

Read 2 more answers

ana and christion collect stamps.together they have 500 stamps in their collection. ana has 150 fewer stamps then christian.how

mezya [45]

Lets represent Ana and Christion using the Letters A and C
a+c=500
c=a+150
now substitute c for a+150 back into the first equation
a+a+150=500
a+a=350
2a=350
Divide by two
a=175
Now that we know annie has 175 all we do is subtract that from the total (500)
C=325
Sure enough, 175 is 150 less than 325
Hope that helped, send me a message if you need clearing up :D

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2 years ago

Which of the following is a radical equation? x + StartRoot 5 EndRoot = 12 x squared = 16 3 + x StartRoot 7 EndRoot = 13 7 Start

d1i1m1o1n [39]

Answer:

x = 3

Step-by-step explanation:

$(8x-8)^{3/2}=64$

Multiply both sides by the exponent 2/3.

$8x - 8 = 64^{2/3}$

Solve for the exponent.

$8x-8=16$

Add 8 to both sides.

$8x = 16+8$

$8x=24$

Divide 8 into both sides.

$x=24/8$

$x=3$

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