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sineoko [7]
2 years ago
6

X+3y=6 Solve for x and y

Mathematics
2 answers:
seropon [69]2 years ago
6 0

Answer:

x=(6,0)

y=(0,2)

or

x=6-3y

y=2-x/3

Step-by-step explanation:

trapecia [35]2 years ago
3 0

Answer:

x=6 and y = 0

Step-by-step explanation:

6+3(0)=6

y = 1

x- 3*1=6

x - 3=6

x = 6+3

x = 9

(x,y)=(9,1)

Let y =0

x - 3*0=6

x - 0=6

x = 6

(x,y) =(6, 0)

Hence, x =6 and y = 0

[RevyBreeze]

You might be interested in
What is 3/4 = w + 2/5
shusha [124]
Convert all fractions to have the same denominator. 

To convert, we must find the least common multiple of the two denominators of the fractions:

Multiples of 4: {4,8,12,16,20}
Multiples of 5: {5,10,15,20}

The least common multiple of 4 and 5 is 20. 

Multiply the fractions by a number that will result in the denominator being 20:

\frac{3}{4} * \frac{5}{5} = \frac{15}{20}

\frac{2}{5} * \frac{4}{4} = \frac{8}{20}

The equation should now look like this:

\frac{15}{20} = w + \frac{8}{20}

Subtract both sides by 8/20 to get w by itself:

\frac{15}{20} - \frac{8}{20} = \frac{7}{20}

w = \frac{7}{20}

The value of w is 7/20.
7 0
3 years ago
Read 2 more answers
Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).​
Reil [10]

Answer:

Part A) sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}

Part B) tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}

Part C) sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}

Step-by-step explanation:

Part A) Find sin(\alpha)\ and\ cos(\beta)

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

sin(\alpha)=cos(\beta)

Find the value of sin(\alpha) in the right triangle of the figure

sin(\alpha)=\frac{8}{14} ---> opposite side divided by the hypotenuse

simplify

sin(\alpha)=\frac{4}{7}

therefore

sin(\alpha)=\frac{4}{7}

cos(\beta)=\frac{4}{7}

Part B) Find tan(\alpha)\ and\ cot(\beta)

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

tan(\alpha)=cot(\beta)

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132

x=\sqrt{132}\ units

simplify

x=2\sqrt{33}\ units

Find the value of tan(\alpha) in the right triangle of the figure

tan(\alpha)=\frac{8}{2\sqrt{33}} ---> opposite side divided by the adjacent side angle alpha

simplify

tan(\alpha)=\frac{4}{\sqrt{33}}

therefore

tan(\alpha)=\frac{4}{\sqrt{33}}

tan(\beta)=\frac{4}{\sqrt{33}}

Part C) Find sec(\alpha)\ and\ csc(\beta)

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

\alpha+\beta=90^o ---> by complementary angles

so

sec(\alpha)=csc(\beta)

Find the value of sec(\alpha) in the right triangle of the figure

sec(\alpha)=\frac{1}{cos(\alpha)}

Find the value of cos(\alpha)

cos(\alpha)=\frac{2\sqrt{33}}{14} ---> adjacent side divided by the hypotenuse

simplify

cos(\alpha)=\frac{\sqrt{33}}{7}

therefore

sec(\alpha)=\frac{7}{\sqrt{33}}

csc(\beta)=\frac{7}{\sqrt{33}}

6 0
3 years ago
The grounds keeper of Estofer Park has to mow 14,623 sq. ft. of grass. He can mow 13 sq. ft of grass in one minute. How many min
MA_775_DIABLO [31]

Answer: 1,124\frac{11}{13}\ minutes

Step-by-step explanation:

Let be "x" the time in minutes that will take him to mow the entire park.

You know that he can mow 13 square feet in 1 minute and he has to mow 14,623 square feet.

Then, you can write the following proportion:

\frac{1}{13}=\frac{x}{14,623}

Solving for "x", you get:

x=\frac{14,623}{13}\ min

To write that improper fraction as a mixed number:

- The quotient when you divide the numerator by the denominator of the fraction obtained, is 1,124 (This will be the whole number).

- The remainder of that division is 11  (This will be the numerator of the fraction).

- The denominator does not change.

Then:

x=1,124\frac{11}{13}\ min

8 0
2 years ago
PLEASE HELP ME I REALLY NEED HELP!!!!!
salantis [7]
<h3>Answer:</h3>

A) x > 2; y < 2x

B) compare the locations of B and C to the shaded region of part A

C) plot the solution space on the graph of schools. points D and E are schools Lisa may attend

<h3>Step-by-step explanation:</h3>

<em>Coordinates and Plotting Points</em>

First of all, you must understand how to plot a point on a graph. Each set of coordinates is an ordered pair. "Pair" means there are two of them. "Ordered" means the sequence in which they appear has significance.

The first number in the pair is the x-coordinate, the distance to the <em>right</em> of the point x=0. (A negative value for this coordinate indicates the distance is to the <em>left</em>.)

The second number in the pair is the y-coordinate, the distance <em>up</em> from the point y=0. (A negative value for this coordinate indicates the distance is <em>down</em>.)

The coordinates (1, 3) for point A mean the point is plotted on the graph 1 unit to the right of x=0 and 3 units up from y=0. The other points are plotted in the same way. (See the labeled points on the attachment, and note their relationship to the horizontal (x) and vertical (y) scales.)

<em>Lines and their equations</em>

A line on a graph is the plot of all the (x, y) pairs on the graph that will satisfy a particular equation. Generally, there will be an infinite number of points—too many to list. For example, some of the points that will satisfy the equation

  y = 2x

are (x, y) = any of ... (0, 0), (0.001, 0.002), (903, 1806), (1.23, 2.46). (Note that the second number in the pair (y) is 2 times the first number (x). That's what y=2x means.) When we want to see the relationship between x-values and y-values that satisfy this equation, it is convenient to plot the line on a graph.

We talk about such lines using terms that describe the steepness of the line (its <em>slope</em>) and whether it goes up to the right (positive slope), down to the right (negative slope), or is vertical (undefined slope) or horizontal (zero slope). In Algebra, as in Geometry, knowing only 2 points is sufficient to define the line we're concerned with. One of the points commonly used to describe a line is its "y-intercept", the point where it crosses the vertical line at x=0.

<em>Inequalities</em>

For problems such as this one, it is sometimes convenient to talk about all the points that are above or below (or left or right) of a given line. The symbols >, <, ≥, and ≤ are used in place of the equal sign in the relation describing these points. A relation that uses one of these symbols instead of the equal sign is called an "inequality."

If we write y < 2x, for example, we mean all the points such that the value of y is less than two times the value of x. Above, we said the point (x, y) = (1.23, 2.46) is on the line y=2x. Now, we can add that the point (1.23, 2.00) is part of the solution to y < 2x, since 2.00 is less than 2×1.23.

Half of the x-y coordinate plane will satisfy any such inequality. We show which half that is by using shading—coloring the portion of the plane where the points meet the requirements of the inequality (are part of the solution). In the attached graph, the solution to y < 2x is colored green. The line at the boundary of that region is dashed because points on that line are <em>equal to 2x</em>. They do <em>not</em> satisfy the requirement that the points be less than 2x.

Part A

For separating a pair of points from the rest of the group, it can be useful to consider where the other points fall in relation to a line through those two points. Here, points B and C are both on the vertical line x=3, and all the other points are to the left of that line.

We also notice that nearby points A and F are on a line with slope somewhere between 1 and 3. That is, a line with slope 2 might be used to separate points A, F (and those to their upper left) from points B and C.

These observations give us ideas for inequalities we can write that separate points B and C from the rest of the group. Two of them might be ...

  • x > 2 (shaded blue)
  • y < 2x (shaded green)

While either of these alone would serve to contain only points B and C, the two of them together form a "system" of inequalities whose solution (overlapping regions) contains only points B and C—as required by the problem statement.

Part B

Points B and C can be verified as solutions by noting their position on the graph relative to the solution regions of the given inequalities.

Part C

The graph also shows a plot of the solution to the inequality y > 2x+5. This inequality also has a boundary line with a slope of 2, and it has a y-intercept of 5 (the point where it crosses the vertical line at x=0). The solution region is shaded red.

We can identify the schools Lisa may attend by their labels in the solution region of the inequality on the graph. Those schools are D and E.

6 0
2 years ago
Could you please help me with hard a problem?
nekit [7.7K]

Answer:123

Step-by-step explanation: plz help me answer my questions plz plz

3 0
3 years ago
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