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2 years ago

Explain why you should record a 1 in the tens column when you regroup in an addition problem

Mathematics

2 answers:

Strike441 [17]2 years ago

6 0

Answer:

Explained below.

Step-by-step explanation:

Consider the addition below:

+ 29

Since the sum in the one's column is 18, a two-digit number, we separate the number as 1 ten and 8 ones.

We now, keep the unit digit 8 and carry over the 1 ten to the tenth column.

Hope this makes sense.

MatroZZZ [7]2 years ago

3 0

Because if u dont regroup u get a completely different answer
19
+19
--------
218

thats what u get if u dont regroup unlike when its done the right way

19
+19
-------
38

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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.

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2 years ago

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The daily consumption of milk in a house is 3.25 litres. How much milk will be consumed in 30 days?

leonid [27]

Answer:

Milk consumed in 30 days = (30 days)(3.25 litres/day) = 97.5 litres

Step-by-step explanation:

You are given a rate at which the milk is consumed (3.25 litres per 1 day). This is just multiplied by the 30 days to get your answer.

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Matt Adams is the best player in his college baseball team. In his last few games, Matt has hit 70%, 77% 81% and 88% of balls th

Ksivusya [100]

79 is the answer ........

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Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation?

Nady [450]

For this case we have the following polynomial:
$x ^ 2 = 12x - 15
$
The first thing to do is to place the variables on the same side of the equation.
We have then:
$x ^ 2 - 12x = -15
$
We complete the square by adding the term (b / 2) ^ 2 on both sides of the equation.
We have then:
$x ^ 2 - 12x + (-12/2) ^ 2 = -15 + (-12/2) ^ 2
$
Rewriting we have:
$x ^ 2 - 12x + (-6) ^ 2 = -15 + (-6) ^ 2

x ^ 2 - 12x + 36 = -15 + 36
 
(x-6) ^ 2 = 21$
$x-6 =+/- \sqrt{21}$
Therefore, the solutions are:
$x = 6 - \sqrt{21}$
$x = 6 + \sqrt{21}$
Answer:
the solution set of the equation is:
$x = 6 - \sqrt{21}$
$x = 6 + \sqrt{21}$

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A line contains the points (82, −96) and (87, −86).

Alik [6]

Answer:

<h2>The answer is 2</h2>

Step-by-step explanation:

The slope of a line given two points can be found by using the formula

$m = \frac{ y_2 - y _ 1}{x_ 2 - x_ 1} \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(82, −96) and (87, −86)

We have

$m = \frac{ - 86 - - 96}{87 - 82} = \frac{ - 86 + 96}{5} = \frac{10}{5} = 2 \\$

We have the final answer as

Hope this helps you

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