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katen-ka-za [31]

2 years ago

12

Consider the graph of Bx + Cy = A where A, B, and C are all positive constants. Find the coordinates of the x-intercept. Use upp

er case letters.

2 answers:

egoroff_w [7]2 years ago

6 0

<h2>Answer:</h2>

The coordinates of the x-intercept is:

$(\dfrac{A}{B},0)$

<h2>Step-by-step explanation:</h2>

The x-intercept of a graph are the points where the graph intersects the x-axis i.e. the points where the y-coordinate is equal to 0.

Here we have the equation of line as:

$Bx+Cy=A$

Now, when y=0 we have:

$Bx=A\\\\i.e.\\\\x=\dfrac{A}{B}$

Hence, the coordinate of the x-intercept is:

$(\dfrac{A}{B},0)$

timofeeve [1]2 years ago

3 0

That is where the line crosses the x axis or where y=0
set y=0
Bx=A
find x value
divie both sides by B
x=A/B
the x intercept is x=A/B

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The angle \theta_1θ

enyata [817]

Answer:

$sin\theta_1 = \dfrac{\sqrt{217}}{19}$

Step-by-step explanation:

It is given that:

$cos\theta_1 = -\dfrac{12}{19}$

And we have to find the value of $sin\theta_1 = ?$

As per trigonometric identities, the relation between $sin\theta\ and \ cos\theta$ can represented as:

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

Putting $\theta_1$ in place of $\theta$ Because we are given

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

Putting value of cosine:

$cos\theta_1 = -\dfrac{12}{19}$

$sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}$

It is given that $\theta_1$ is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

$\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}$

8 0

2 years ago

The projectile motion of an object can be modeled using s(t) = gt2 + v0t + s0, where g is the acceleration due to gravity, t is

NeX [460]

S(t) = -4.9t^2 + 19.6t + 24.5

when it reaches the ground s(t) = 0

so we have -4.9t^2 + 19.6t + 24.5 = 0

solve this for t:-

-4.9(t^2 - 4t - 5) = 0

(t + 1)(t - 5) = 0

so required time t = 5 seconds

answer 5 seconds

5 0

2 years ago

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The championship record in baseball in a certain year was 52 games won, and 14 lost. What percent of the games were won?

slamgirl [31]

Number of games won/Number of total games played x 100%

52/66 x 100% = 78.79%

Hope u understood! :)

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2 years ago

Monica has a bag of marbles. There are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick

natima [27]

Answer:

$\frac{7}{64}\approx 0.11$

Step-by-step explanation:

We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.

Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.

Since the probability of two independent compound events is always the product of probabilities of both events.

$P(\text{A and B})=P(A)*P(B)$

Now let us find probability of picking a red marble out of 16 (4+7+5) marbles.

$P(Red)=\frac{\text{Total red marbles}}{\text{Total marbles}}$

$P(Red)=\frac{4}{16}$

Probability of picking blue ball out of 16 (4+7+5) marbles:

$P(Blue)=\frac{\text{Total blue marbles}}{\text{Total marbles}}$

$P(Blue)=\frac{7}{16}$

Now let us find probability of Monica picking a red and then a blue marble.

$P(\text{Red and Blue})=\frac{4}{16}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{1}{4}\times \frac{7}{16}$

$P(\text{Red and Blue})=\frac{7}{4*16}$

$P(\text{Red and Blue})=\frac{7}{64}$

$P(\text{Red and Blue})=0.109375\approx 0.11$

Therefore, the probability of picking a red and then blue marble is $\frac{7}{64}\approx 0.11$ .

7 0

2 years ago

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Mrac [35]

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3 0

2 years ago

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