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2 years ago

Factor

2 answers:

6 0

The answer would be...

6(x-1)(2x^4+2x^3+2x^2+3x+3)

I hope this helped but I couldn't see this as one of the multiple choice answers but I worked it out and this I what I got

stellarik [79]2 years ago

5 0

C on that one which is the correct answer

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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:

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

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

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

Javier has four cylindrical models. The heights, radii, and diagonals of the vertical cross-sections of the models are shown in

Sever21 [200]

Given the heights, radii, and diagonals of the vertical cross-sections of the models, the model in which the lateral surface meet the base at a right angle is the model in which the height, the diameter and the diagonal of the vertical cross-section forms a right triangle.

i.e. the sum of the squares of the height (h) and the diameter (d) gives the square of the diagonal vertical cross-section (l).

For model 1:

radius: 14 cm, thus diameter = 2(14) = 28 cm
height: 48 cm
diagonal: 50 cm

 $d^2+h^2=28^2+48^2 \\ \\ =784+2,304=3,090\neq50^2=l^2$

Thus, the lateral surface of model 1 does not meets the base at right angle.

For model 2:

radius: 6 cm, thus diameter = 2(6) = 12 cm
height: 35 cm
diagonal: 37 cm

[tex]d^2+h^2=12^2+35^2 \\ \\ =144+1,225=1,369=37^2=l^2[/tex]

Thus, the lateral surface of model 2 meets the base at right angle.

For model 3:

radius: 20 cm, thus, diameter = 2(20) = 40 cm
height: 40 cm
diagonal: 60 cm

 $d^2+h^2=40^2+40^2 \\ \\ =1,600+1,600=3,200\neq60^2=l^2$

Thus, the lateral surface of model 3 does not meets the base at right angle.

For model 4:

radius: 24 cm, thus, diameter = 2(24) = 48 cm
height: 9 cm
diagonal: 30 cm

 $d^2+h^2=48^2+9^2 \\ \\ =2,304+81=2,385\neq30^2=l^2$

Thus, the lateral surface of model 3 does not meets the base at right angle.

Therefore, the model in which the lateral surface meets the base at a right angle is model 2 (option b)

8 0

2 years ago

Read 2 more answers

((whoever gets this right gets brainliest))

Papessa [141]

Answer:

The answer would be A because if you look at the G then the H and then the E, Look at it from the exact same order expect from the left.

I really hope my answer will help you!

Good Luck!

3 0

2 years ago

Read 2 more answers

The scatter plot shows a scuba diver's depth in the ocean. The equation of the trend link shown is y = -3.29x - 10. Predict what

TiliK225 [7]

It's not given a scatter plot not the characteristics of the variables, but it can be safely assumed as shown below

Answer:

The diver's depth will be -42.9 after 30 minutes

Step-by-step explanation:

The equation of the trend (or best fit line) for the scuba diver's depth in the ocean is:

y = -3.29x - 10

Where y is the diver's depth and x is the time in minutes.

To predict the diver's depth at x=30 minutes, substitute in the equation:

y = -3.29(10) - 10

y = -32.9 - 10

y = -42.9

Since no units are provided for y, the answer is:

The diver's depth will be -42.9 after 30 minutes

6 0

2 years ago

An empty can weighs 30 grams. The weight of the same can filled with liquid is 47.3 grams. Which equation could be used to find

Alika [10]

Answer:

w = y - x is the equation to find w

Where,

w = Weight of the liquid

x = Weight of the empty can

y = Weight of the can with liquid

Step-by-step explanation:

Let

Weight of the empty can = x

Weight of the liquid = w

Weight of the can with liquid = x + w = y

x= 30 grams

w = ?

y= 47.3 grams

y = x + w

w = y - x

= 47.3 - 30

= 17.3

w= 17.3 grams

Therefore, the weight of the liquid is 17 .3 grams

3 0

2 years ago