Michael invests $2000 in an annuity that offers an interest rate of 4%

A line that has a rise of -34 and a run of 20.

aleksandr82 [10.1K]

Answer:

$y=-1.7x-34$

Step-by-step explanation:

An equation of a line is given $y=mx+c$ , where m is the gradient or $\frac{Rise}{Run}$

The rise indicates the change in value of y (vertical change in the graph).

The run indicates the change in value of x (horizontal change in the graph).

Let point 1 of the graph be (x,y) = (0,-34) and point 2 be (x,y) = (20,0).

$m =\frac{Rise}{Run} \\ =\frac{-34}{20}\\ =-1.7$

Substitute point 1 into the graph to get the constant c.

$y=mx+c$

$-34=(-1.7)*(0)+c$

$c=-34$

∴ The equation of the line is, $y=-1.7x-34$

7 0

3 years ago

Roxanne's car used 4.8 gallons of gasoline to drive 124 miles. If Roxanne has 180 more miles to go, which is closest to the addi

Reika [66]

She would use 6 i believe

3 0

3 years ago

Weekly wages at a certain factory

shtirl [24]

Where is the question

6 0

3 years ago

HELP ASAP!!! Please give a step-by-step explanation, or you will be reported!

stepan [7]

For all the dogs: you have 2.65+2.46+3.67+2.91 = 11.69%

Now, the probability of getting a chihuahua would be 3.44/11.69=0.2942= 24.92

Step-by-step explanation:

there ya go!!

6 0

3 years ago

Read 2 more answers

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago