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3 years ago

Which factor is important to consider when building a saving fund for planned long term needs ?

1 answer:

5 0

Answer: There are several factors to consider when building a savings fund for planned long term needs. First of all, we have to consider the amount of principal money we can invest in. It may be a one-time investment or a systematic investment plan.

Step-by-step explanation: Indeed, this is from Brainly so this is NOT my answer. Just gave it to you to help you out. Good lucky, my friend.

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Consider the parabola r(t)equalsleft angle at squared plus 1 comma t right angle, for minusinfinityless thantless thaninfinity

kodGreya [7K]

Given:- $r(t)=< at^2+1,t>$ ; $-\infty < t< \infty$ , where a is any positive real number.

Consider the helix parabolic equation :

$r(t)=< at^2+1,t>$

now, take the derivatives we get;

$r{}'(t)=$

As, we know that two vectors are orthogonal if their dot product is zero.

Here, $r(t) and r{}'(t)$ are orthogonal i.e, $r\cdot r{}'=0$

Therefore, we have ,

$< at^2+1,t>\cdot < 2at,1>=0$

$< at^2+1,t>\cdot < 2at,1>=$

$=2a^2t^3+2at+t$

$2a^2t^3+2at+t=0$

take t common in above equation we get,

$t\cdot \left (2a^2t^2+2a+1\right )=0$

⇒ $t=0$ or $2a^2t^2+2a+1=0$

To find the solution for t;

take $2a^2t^2+2a+1=0$

The number $D = b^2 -4ac$ determined from the coefficients of the equation $ax^2 + bx + c = 0.$

The determinant $D=0-4(2a^2)(2a+1)$ $=-8a^2\cdot(2a+1)$

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola $r(t)=< at^2+1,t>$ i.e <1,0>

6 0

3 years ago

Easy Multiple Choice Questions about expressions and linear equations / ASAP / Will mark brainliest / 25 Points

nata0808 [166]

Answer:

1.C

2.A

3.D

4.C

5.B

6.A

7.B

8.C

9.A

10. D

Step-by-step explanation:

For Problem 1, the work is as follows:

$2(3x-1)+x\\Distribute\\6x-2+x\\7x-2$

For Problem 2, the work is as follows:

$(3x^{2} +6x+4) - (-x^{2} +2x-1)\\Distribute -\\(3x^{2} +6x+4) + (x^{2} -2x+1)\\Combine\\4x^{2} +4x+5$

For Problem 3, the work is as follows:

$5b-(a+c)^{2} \\5(3) - (-1+-4)^{2} \\15 - (-5)^{2} \\15-25\\-10$

For Problem 4, the work is as follows:

$4x-8 = -12\\+8\\4x = -4\\-4/4 = -1\\x = -1$

For Problem 5, the work is as follows:

$\frac{1}{2} (12x-6) = 2x-15\\Distribute\\\\6x - 3 = 2x -15\\+3\\6x = 2x -12\\-2x\\4x = -12\\-12/4\\x = -3$

For Problem 6, the work is as follows:

$17 *0.06 = 1.02\\17+1.02 = 18.02$

For Problem 7, the work is as follows:

$x = (2,0)\\y = (0,-1)\\\frac{rise}{run} \\y = \frac{1}{2} x-1$

For Problem 8, the work is as follows:

$y_{2} -y_{1} \\x_{2} -x_{1} \\-1-1 = -2\\0-(-2) = 2\\Slope = -1$

For Problem 9, the work is as follows:

$3y - 2x +6 = 0\\-3y\\(-3y - -2x+6)/-3\\y = \frac{2}{3} x -2$

For Problem 10, the work is as follows:

$(y-3) = -2(x-1)\\Distribute\\(y-3) = -2x+2\\+3\\y = -2x +5$

Hope this Helps!

4 0

2 years ago

Please help asap!!!! 53 points!

AfilCa [17]

The easiest way is to graph it based upon the slope (m) and y-intercept (b), in the standard slope-intercept form: y = m (x) + b.
The line above intercepts the y-axis at y = -2, which is b. The slope (m) = rise/run = (y2-y1)/(x2-x1 ); so for the point (-4, 2) to (-6, 4) is:
(4-2)/(-6--4) = 2/(-6+4) = 2/-2 = -1.
So one form of the equation would be:
y = -1x - 2
Now the other form of an equation is point-slope: y-k = m (x-h), where the point is at (h, k)
and if we pick -5 for x (bc 5 it listed in 3 of the answers), the y at x=-5 looks like around +3
so we get: y-k = -1 (x--5)...
y-3 = -(x+5)... therefore D) is the correct answer:

8 0

3 years ago

Read 2 more answers

The bones of a mammal were discovered to have 3% as much Carbon-14 as they would have contained when

tankabanditka [31]

I be flexing 3 dollars carbon-144

5 0

3 years ago

Almost there .......AAAAHHhH

madreJ [45]

Answer:

12/46