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2 years ago

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Part A The mosquito population for the Eastern United States was predicted to be 23,467,899,213. Which number estimates the pred

icted mosquito population?
2 × 10 9

2 × 10 ¯ 10

2 × 10 10

2 × 10 ¯ 11

1 answer:

irina1246 [14]2 years ago

6 0

Answer:

2x10 in power 10

Step-by-step explanation:

23467899213 estimate 20000000000, 2x 10000000000=2x10^10

You might be interested in

Which expression(s) have a greatest common factor (GCF) of 3xy2 with 42xy4

Aleks [24]

Answer:

3xy2

Step-by-step explanation:

GCF is the largest factors which are common in two or more numbers.

We find the GCF of two or more number number by breaking each number into their factor and then observe which factors are common in the group and then we multiple the common number to get the GCF.

Lets understand this by an example

GCF of 24 and 36

factors of 12 = 2*2*2*3 , factors of 36 = 2*2*3*3

hence we see that 2,2 and 3 are common in 12 and 36 hence

GCF will be 2*2*3 = 12

Coming back to problem

given

Expression 1 is 3xy^2

factor of 3xy^2 = 3*x*y*y

Expression 2 is 42xy4

factor of 42xy4 = 2*3*7*x*y*y*y*y

common factor between the two expressions are

3, x, y , y

hence GCF will be = 3*x*y*y =3xy2 Answer

3 0

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

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 31% of courses have a res

beks73 [17]

Answer:

0.25

Step-by-step explanation:

72% of courses have final exams and 46% of courses require research papers which means probability of 0.72 for courses that have final exams and 0.46 for courses that require research papers.

31% of courses have a research paper and a final exam, which means probability of 0.31 for both courses with exams and research papers, using Venn diagram approach, find picture attached to the solution.

P(R or E) = P(R) + P(E) - P(R and E), which gives:

P(R or E) = 0.15 + 0.41 - 0.31

P(R or E) = 0.25.

6 0

2 years ago

(01.02 LC)Simplify 9 − (−4).

levacccp [35]

Answer:

13

Step-by-step explanation:

When we subtract a negative, it is like adding

9 - -4

9 +4

13

3 0

2 years ago

Read 2 more answers

Prescription for Marilyn Jones

Vanyuwa [196]

Answer:

32

Step-by-step explanation:

We can ignore Piroxicam, because it comes in capsules.

1. Number of Glucotrol tablets

Marilyn takes one tablet per day.

$\text{No. of tablets} = \text{8 da} \times \dfrac{\text{1 tab}}{\text{1 da}} = \text{8 tab}$

2. Number of Isoniazid tablets

Marilyn takes three tablets per day.

$\text{No. tablets} = \text{8 da} \times \dfrac{\text{3 tab}}{\text{1 da}} =\text{24 tab}$

3. Total tablets

Total tablets = 8 tab + 24 tab = 32 tab

Marilyn takes 32 tablets in eight days.

7 0

2 years ago