What is the value of X?

Plz help.......................

slamgirl [31]

The second graph, B is the answer

8 0

4 years ago

1. Rewrite this relation as a function with the dependent variable on the left side of the equation. S+t2=16

Sauron [17]

Answer: d. s=16-t²

Step-by-step explanation: the answer is s=16-t², because for this case is really important to keep in mind which is the value with the quadratic expression, when we have a quadratic expression the graphic representation will be a parabola, in a parabola you will find a point that never change which is constant in the diagram.

also keep in mind that the values which are placed in the axis of the ordinates are the independent variables and in parabola the quadratic values are placed in the axis of the ordinates , and for this case is t² because of the quadratic expression, then we can find that t² is the independent variable and s is the dependent variable.

3 0

3 years ago

Tim Jones bought 100 shares of a mutual fund ABC at $5.25 with no load. And 100 shares of DEF at $6 which had a load of $275.

ipn [44]

Question Completion:

Evaluate the two investments, assuming shares of ABC were sold at $10.00 each, while those of DEF were sold at $15 each. Which investments yielded the higher return?

Answer:

Tim Jones

Investment in DEF yielded the higher return of $1,225 (net proceeds) than ABC's return of $1,000.

Step-by-step explanation:

a) Data and Calculations:

Investments in ABC and DEF:

ABC DEF

Investment (shares) 100 100

Unit price of shares $5.25 $6.00

Total investments $525 $600

Selling price per share $10 $15

Sales proceeds $1,000 $1,500

Investment commission ($0) ($275)

Net proceeds from sale $1,000 $1,225

b) The above calculations show clearly that the investment in DEF with a load of $275 yielded the higher returns than investment in ABC with no load. This result suggests that the investment load may not be the only factor for rejecting an investment, but its net proceeds.

4 0

3 years ago

Read 2 more answers

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$