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3 years ago

Sing white and red paint, which two mixtures result in the same shade of pink?

Mathematics

2 answers:

nadya68 [22]3 years ago

7 0

B and c is the answer

Bumek [7]3 years ago

5 0

Answer:

B & C

Step-by-step explanation:

Mixture B:

4 white 2 red

Mixture C:

Divide the whole paint combinations by 2:

2 white 1 red

In that case, you are simply making less, but still have the same ratios. They are proportionate.

Happy learning!

--Applepi101

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A father is 51 years old and his son is 19. How many years ago was the father 5 times his son's age?

Jet001 [13]

Answer:

x=11

Step-by-step explanation:

51-x-5*19= -5x

51-5*19= -4x

4x=5*19-51

4x=95-51

4x=44

x=11

5 0

3 years ago

Read 2 more answers

The equation a = StartFraction one-half EndFraction left-parenthesis b 1 plus b 1 right-parenthesis.(b1 + b2)h can be used to de

aleksley [76]

Answer:

The equivalent expressions are:

$b1=\frac{2a}{h}-b2$

$h=\frac{2a}{b1+b2}$

Step-by-step explanation:

Given equation for finding area of a trapezoid:

$a=\frac{1}{2}(b1+b2)h\\$

where $a$ represents area, $h$ represents height and $b1\ and\ b2$ represents the base lengths of the trapezoid.

Evaluating $h$ by rearranging the equation to find an equivalent equation.

Multiplying both sides by 2.

$2\times a=2\times\frac{1}{2}(b1+b2)h$

$2a=(b1+b2)h$

Dividing both sides by $b1+b2$

$\frac{2a}{b1+b2}=\frac{(b1+b2)h}{b1+b2}$

$\frac{2a}{b1+b2}=h$

$\therefore h=\frac{2a}{b1+b2}$

Evaluating $b1$ by rearranging the equation to find an equivalent equation.

Multiplying both sides by 2.

$2\times a=2\times\frac{1}{2}(b1+b2)h$

$2a=(b1+b2)h$

Dividing both sides by $h$

$\frac{2a}{h}=\frac{(b1+b2)h}{h}$

$\frac{2a}{h}=b1+b2$

Subtracting both sides by $b2$

$\frac{2a}{h}-b2=b1+b2-b2$

$\frac{2a}{h}-b2=b1$

$\therefore b1=\frac{2a}{h}-b2$

8 0

3 years ago

Read 2 more answers

Can someone please help

atroni [7]

Answer:

Step-by-step explanation:

$\frac{5}{24}+\frac{2}{3-x}=\frac{1}{4}\\\\\frac{5}{24}-\frac{1}{4}=\frac{-2}{3-x}\\\\\frac{5}{24}-\frac{1*6}{4*6}=\frac{-2}{3-x}\\\\\frac{5-6}{24}}=\frac{-2}{3-x}\\\\\frac{-1}{24}}=\frac{-2}{3-x}\\\\\\(-1)*(3-x)=(-2)*24\\\\-1*3-(-1)*x = -48\\\\-3 + x = -48\\\\x = -48 + 3\\\\x = -45$

x = -45

4 0

3 years ago

A system of equations is created by using the line represented by 2 x 4 y = 0 and the line represented by the data in the table

Marta_Voda [28]

Solution to a system of equation is values of variables true for all equations of that system.The x-value of solution to the given system is 2

<h3>How to find the solution to the given system of equation?</h3>

For that , we will try solving it first using the method of substitution in which we express one variable in other variable's form and then you can substitute this value in other equation to get linear equation in one variable.

If there comes a = a situation for any a, then there are infinite solutions.

If there comes wrong equality, say for example, 3=2, then there are no solutions, else there is one unique solution to the given system of equations.

<h3>What is the equation of a line passing through two given points in 2 dimensional plane?</h3>

Suppose the given points are $(x_1, y_1)$ and $(x_2, y_2)$ , the n the equation of the straight line joining both two points is given by

$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1} (x -x_1)$

The first equation is $2x + 4y = 0$

Since the second equation of line contains points given, let we consider two points,

$(x_1, y_1) = (-1,8)\\(x_2, y_2) = (3, -4)$

Then, the equation of line is found as:

$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1} (x -x_1)\\\\(y - 8) = \dfrac{-4-8}{3-(-1)}(x - (-1))\\\\y = -3(x + 1) + 8\\\\3x + y = 5$

Thus, the system of equations is

$2x + 4y = 0\\3x + y = 5$

From second equation, getting y in terms of x(since its easy as y has 1 as coefficient), we get:

$3x + y = 5\\y = 5 - 3x$

Putting this value of y in first equation, we get:

$2x + 4y = 0\\\\2x + 4(5-3x) = 0\\20 = 10x\\x= 2$

Putting this value of x in expression for y, we get:

$y = 5 - 3x = 5 - 6 = -1$

Thus, the solution to the system of equations obtained is (x,y) = (2,-1)

Thus, The x-value of solution to the given system is 2

Learn more about system of equations here:

brainly.com/question/13722693

4 0

3 years ago

The center of an ellipse is located at (3, 2) One focus is located at (6, 2) and its associated directrix is represented by the

tangare [24]

Answer:

Step-by-step explanation:

This is really hard to try and explain over the internet here, but I'll do my best, assuming you have some experience with ellipses.

We are given the center of (3, 2), and the focus of (6, 2). The focus is 3 units to the right of the center; the other focus is 3 units to the left of the center at (0, 2). The focus has coordinates of (-ae, 2) and (ae, 2) from left to right and the distance between them then is 2ae. There are 6 units between the 2 foci, so 2ae = 6.

The directrix is 8 1/3 units from the center to the right AND the left. The directrices have distances of 2a/e between them. There are 16 2/3 units between the 2 directrices, so

$\frac{2a}{e}=\frac{50}{3}$

We can solve for the eccentricity of this ellipse given the fact that the eccentricity is the ratio of the distance between the foci to the distance between the directrices. Therefore,

$\frac{2ae}{2\frac{a}{e} }=\frac{6}{\frac{50}{3} }$

The 2's andd the a's cancel, leaving you with

$e^2=\frac{18}{50}=\frac{9}{25}$

That gives us then that

$e=\frac{3}{5}$

Going back to the identity for the distance between the foci, 2ae = 6, we fill in e and solve for a:

$2a(\frac{3}{5})=6$ and

$\frac{6a}{5}=6$

which gives us that a = 5

Now use the identity $b^2=a^2(1-e^2)$ to solve for b:

$b^2=25(1-\frac{9}{25})$ and

$b^2=25(\frac{16}{25})$ and

b = 4

Because this is a horizontally stretched ellipse, it is of the form

$\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1$

we fill in as

$\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$

Again, since you are expected to be able to solve for the equation of an ellipse at this advanced level, I am assuming that the equations I gave above as a means to solve for the different characteristics of an ellipse are familiar to you. This is definitely NOT beginning conics!

3 0

3 years ago

Read 2 more answers