All categories

3 years ago

Panel 1 is the answer choice Panel 2 is the question

Mathematics

1 answer:

Ierofanga [76]3 years ago

8 0

D is the midpoint of A and B
A = (0,s)
B = (r,0)

Add up the x coordinates to get 0+r = r. Then cut that in half to get r/2
Similarly, do the same for the y coordinates: s+0 = s ---> s/2

So the location of point D is (r/2, s/2)

Answer: Choice C

You might be interested in

The triangles shown below must be congruent. right awnsers only please.

tankabanditka [31]

It is true. The triangles are congruent.

8 0

3 years ago

Read 2 more answers

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago

How many of the first 1000 positive integers are multiples of both 4 and 5 but not 6 ?

castortr0y [4]

Let's begin by breaking each number down into its prime factors: 4 = 2 x 2 5 = 5 6 = 2 x 3 Next, let's determine the Lowest Common Multiple (LCM) of the numbers 4, 5, and 6 by multiplying all common and unique prime factors of each number: common prime factors: 2 unique prime factors: 2,5,3 LCM = 2 x 2 x 5 x 3 = 60 Next, let's determine how many times 60 goes into 10,000 (excluding remainder): 10,000/60 = 166 and 2/3 Multiples of ALL 3 numbers (4,5,6) = 166 Next, let's determine the Lowest Common Multiple (LCM) of the numbers 4 and 5 by multiplying all common and unique prime factors of each number: common prime factors: none
unique prime factors: 2 x 2 x 5
LCM = 2 x 2 x 5 = 20 Next, let's determine how many times 20 goes into 10,000:
10,000/20 = 500
Multiples of BOTH numbers (4 and 5) = 500 Finally, let's subtract the multiples of ALL three numbers (4,5,6) from the multiples of BOTH numbers (4 and 5) to get our answer: Multiples of ONLY numbers 4 and 5 (excluding 6): 500 - 166 = 334

5 0

3 years ago

What is 20-2w^2 when, w= 3

kow [346]

Answer:

Step-by-step explanation:

Use order of operations PEMDAS

3^2 = 9 * 2 = 18

20-18= 2

7 0

3 years ago

Translate the following word phrase into nine times the difference of 5 and y

emmasim [6.3K]

Answer: 9 * (5 - y)

Explanation:

Times: multiplication

Difference: subtraction

y: the unknown variable

4 0

3 years ago