The graph shows two lines, A and B.

Mathematics

2 answers:

scoundrel [369]3 years ago

8 0

Answer:

second option: (4, 4) is the solution to both lines A and B.

Step-by-step explanation:

You know that the equation of line A is:

$y=-3x+15$

and the equation of line B is:

$y=\frac{1}{2}x+2$

The point in which the line A intersects with the line B is the solution of the sytstem of equations.

You can observe in given graph that the point of intersection of Line A and Line B is: (4,4)

Therefore (4, 4) is the solution to both lines A and B.

Svetach [21]3 years ago

6 0

Answer:

(4, 4) is the solution to both lines A and B.

Step-by-step explanation:

Given graph shows two lines, A and B.

Based on the graph, we need to find about which statement is correct about the solution to the system of equations for lines A and B. Where give choices are:

(4, 4) is the solution to line A but not to line B.

(4, 4) is the solution to both lines A and B.

(5, 1) is the solution to line B but not to line A.

(0, 2) is the solution to both lines A and B.

From graph we can see that both lines intersect at point (4,4). Since solution lies on both lines so the correct choice is:

(4, 4) is the solution to both lines A and B.

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A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$