All categories

3 years ago

Find the slope of a line perpendicular to the line whose equation is 3y + 2x = 6

1 answer:

4 0

$3y + 2x = 6\\ \\the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\ \\3y=-2x+6 \ \ /:3\\ \\y=-\frac{2}{3}x+\frac{6}{3}\\ \\y=-\frac{2}{3}x +2 \\ \\ m = -\frac{2}{3} \\ \\ Answer : \ slope \ is \ m = -\frac{2}{3}$

You might be interested in

Please I need help. you will get points

iragen [17]

Answer:

t(32) = 94

Step-by-step explanation:

a. I think s is the independent variable

b I think t or t(s) is the dependent variable since the total amount of money depends on how many treats are sold.

t(32) = 30 + 2(32)

t(32) = 30 + 64

t(32) = 94

I'm a little unsure on the dependent and independent variable but I know the function is correct.

Hope this helps :)

7 0

3 years ago

Jill ran X Miles yesterday, today she ran 2 miles more than yesterday , with a total greater of 6 miles . Circle the numbers bel

Eddi Din [679]

Five, Six, Seven, and eight

8 0

3 years ago

Read 2 more answers

Write a literal corresponding to the floating point value one-and-a-half.

kotykmax [81]

1.5 In C or C++, a floating point literal is nothing more than a sequence of decimal digits with either or both an decimal point, or an exponent with optionally a type specifier. The value could be any of the following: 1.5, 1.5e0, 0.15e1, 15e-1, etc.

7 0

3 years ago

If c = 4 and d = 5, find c:d. 1:5 4:5 5:4

nadya68 [22]

The answer would be the second choice. 4:5

5 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