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3 years ago

Which set(s) of points show x in DIRECT PROPORTION to y?

Mathematics

2 answers:

hodyreva [135]3 years ago

6 0

Answer:D){F, G, H} and {G, C, A} only

Step-by-step explanation:

vredina [299]3 years ago

5 0

Answer: OPTION D.

Step-by-step explanation:

For this exercise it is important to remember that:

1. Direct proportion equations have the following form:

$y=kx$

Where "k" is the constant of proportionality.

2. The graph of Direct proportions is an straight line that passes through the origin.

Observe the picture attached.

If you join the points G, C and A, you get a straight line that passes through the origin,

If you join the points F, G and H, you also get a straight line that passes through the origin,

Therefore, the sets of points show "x" in Direct proportion to "y", are:

$\{F, G, H\}$ and $\{{G, C, A}\}$

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Solve -1 < x + 1 ≤ 6

andrey2020 [161]

Answer:

-2<x≤5

Step-by-step :

-1<x+1≤6 So fisrt what every we do to one side we must do to the other. In this case it's a bit different since we are dealing with inequalities.

-1<x+1≤6 I would start off by isolating x in the middle.

-1<x+1≤6 I subtracted 1 from all three sides.

-1 -1 -1

Now your equation should look like this:

-2<x≤5 Now there is really nothing much we can do here since we were just trying to get x by its self.

Answer : -2<x≤5

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3 years ago

Helllppp plzzzzzzzzzzz

kolbaska11 [484]

Answer:

A: 5

B: 32

C: 3

D; 14

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Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago

Simplify (7+5)^2 divided by 4 x 3 + 9

MaRussiya [10]

((7 + 5)^2) / (4 * 3 + 9) =
(12^2) / (12 + 9) =
144 / 21 reduces to 48/7 or 6 6/7

7 0

4 years ago

Write the equation for a line in standard form (Ax+By+C) that is perpendicular to y = 3x -

salantis [7]

Answer:

x+3y-6=0

Step-by-step explanation:

given eqn is y=3x-2 which is 3x-y-2=0

the eqn of line perpendicular to given eqn is -x+3y+k=0

it passes through (6,4)

-6+3*4+k=0

or,. -6+12+k=0

or, k= -6

therefore, the eqn of line perpendicular to given eqn is x+3y-6=0

8 0

4 years ago