Can some one help me with this problem

How is the diameter of a circle related to the distance around a circle

Anastasy [175]

The distance around a circle on the other hand is called circumference and half of the diameter or the midpoint th the circle border is called the radius of the circle.

5 0

3 years ago

Which point-slope equation represents a line that passes through (3,-2) with a slope of -4/5?

Viefleur [7K]

Answer:

y + 2 = -4/5 (x - 3)

Step-by-step explanation:

y - y1 = m (x - x1). Put in your points for y1 and x1, and always remember to flip your signs.

unfortunately, I don't know why that isn't one of your answers, so sticking with the positive 2 I would choose y+2=- (- 3) if I were you. Looks like the -4/5 disappeared somehow?

Hope this helps

8 0

3 years ago

Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

3 years ago

there are 13 animals in a barn. some are ducks and some are pigs. there are 40 legs in all. how many of each animal are there?

konstantin123 [22]

Let X be chickens
Let Y be pigs,
then x+y=13
the animals have 2x+4y legs,
so 2x+4y=40
The system of the equations
[X+y=13
[2x+4y=40
Divide the second equation by 2
[x+y=13
[x+2y=20
subtract the first equation from the second one
y=20-13
y-7 pigs
substitute y=7 into the first equation
x+7=13
x=13-7
x=6 chickens

6 0

3 years ago

The coffee counter charges ?"$11.00" per pound for kenyan french roast coffee and ?"$13.00" per pound for sumatra coffee. How mu

zloy xaker [14]

Answer: The answer is 13 ponds of each type should be used.

Step-by-step explanation: Given that the coffee counter charges $11.00 per pound for kenyan french roast coffee and $13.00 per pound for sumatra coffee. We are to find the quantity of each type that should be used to make a 26 pound blend that sells for $12.00 per pound.

Let 'x' pound and 'y' pound of kenyan french roast coffee and sumatra coffee be used in the mixture of 26 pound.

So, we have

$x+y=26,~~~~~~~~~~~~~~~~~~~~~~(A)\\\\11x+13y=12\times 26.~~~~~~~~~~(B)$

Multiplying equation (A) by 11 and subtracting from equation (B), we have

$13y-11y=12\times26-11\times26\\\\\Rightarrow 2y=26\\\\\Rightarrow y=13,$

and from equation (A),

$x=26-13=13.$

Thus, 13 pounds of each type of coffee should be used.

8 0

3 years ago