Choose the set of equivalent fractions that correctly uses the LCD for these fractions: 2/3 and 1/6

Mathematics

1 answer:

andreev551 [17]3 years ago

4 0

Answer:

4/6 and 1/6

Step-by-step explanation:

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Is the point (3, 11) a solution for the linear equation 1/3x + y = 12? Explain.

Brut [27]

Answer:

Yes, because when the values are substituted in, the equation becomes a true statement

Step-by-step explanation:

plug in the values for x and y and see if the equation makes sense

x=3

y=11

1/3(3)+11=12

1+11=12

12=12

Since the above equation is true, the point is a solution for the linear equation.

7 0

3 years ago

Otto used 5.5 cups of whole wheat flour and x cups of white flour in the recipe. What is the equation that can be used to find t

sashaice [31]

Answer:

This situation of the recipe can be represented as y=x+5.5 and it is expected that x is ≥ 0 and y is ≥ 5.5.

What is the equation for this situation?

It is known the total amount of flour (y) is equivalent to the total amount of whole wheat flour (5.5) added to the total white flour (x). Therefore, the equation is:

y = x + 5.5

The possible values are

x = It is expected x is equalthan 0 or greater to it since the minimum amount of white flour that can be added is 0.

y = It is expected y is equal than 5.5 or greater to it since even if no

white flour is added the minimum total is 5.5.

Learn more about equation in:

brainly.com/question/10413253

Step-by-step explanation:

8 0

2 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$