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IgorLugansk [536]
3 years ago
15

Evaluate the variable expression when a=-4, b=2, c=-3, and d =4. b-3a/bc^2-d​

Mathematics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

Step-by-step explanation:

Evaluate:

\dfrac{b-3a}{bc^{2}-d}

When a=-4, b=2, c=-3, and d =4

Solution:

Substitute, a=-4, b=2, c=-3, and d =4 in above expression we get

\dfrac{b-3a}{bc^{2}-d}=\dfrac{2-3(-4)}{2(-3)^{2}-4}\\\\=\dfrac{2+12}{18-4}\\\\

\dfrac{b-3a}{bc^{2}-d}=\dfrac{14}{14}=1

Therefore, the variable expression when a=-4, b=2, c=-3, and d =4 is

\dfrac{b-3a}{bc^{2}-d}=1

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Write an equation of the line that passes through the given point and is parallel to the given line.
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Answer : <em>Equation of line is</em> y=Equation of line is y=\frac{2}{3}x+\frac{-5}{3}

Step-by-step explanation:

Theory :

Equation of line is given as y = mx + c.

Where, m is slope and c is y intercepted.

Slope of given line : y = \frac{2}{3}x+1 is m= \frac{2}{3}

We know that line : y = \frac{2}{3}x+1 is parallel to equation of target line.

therefore, slope of target line will be \frac{2}{3}.

we write equation of target line as y= \frac{2}{3}x+c

Now, It is given that target line passes through point ( -5,-2)

hence, point ( -5,-2) satisfy the target line's equation.

we get,

y= \frac{2}{3}x+c

-2= \frac{2}{3} \times -5+ c

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c= \frac{-5}{3}

thus, Equation of line is y=Equation of line is y=\frac{2}{3}x+\frac{-5}{3}

7 0
3 years ago
Susan and Mark are standing at different places on a beach and watching a bird. The angles of elevation they make are 20º and 50
Mandarinka [93]

Answer:

The bird is 2.44km high

Step-by-step explanation:

Hello,

To solve this question, we need to understand how they are and we can only get this with a correct pictorial diagram.

See attached document for better understanding.

From the first diagram, we understand that the bird is between them and also on top of them.

Assuming Susan, Mark and the bird all form a triangle at each other and the bird at the top, we can divide the the triangle into two equal parts.

But before then, we should know that sum of angles in a triangle is equal to 180°

Therefore,

20° + 50° + b° = 180

70° + b = 180

b = 180° - 70°

b = 110°

Dividing angle b into two equal parts = 55° on each side.

See the last attached document for better illustration.

Using SOHCAHTOA, we can find the adjacent of the triangle which corresponds to the height of the bird.

We have opposite = 3.4km and we're looking for adjacent. We can use tangent of the angle to find the adjacent.

Tanθ = opposite / adjacent

Tan 55° = 3.5 / adj

Adjacent = 3.5 / tan55

Adjacent = 3.5 / 1.43

Adjacent = 2.44km

The height of the bird is 2.44km

7 0
3 years ago
Please help answer thisq​
Julli [10]

Answer:

More than one independent variable being tested.

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5 0
3 years ago
A plane flew 360km in 3 hrs when flying with the wind.With no change in the wind,the return trip took 4 hrs.Find the speed of th
morpeh [17]
Speed of the plane: 250 mph
Speed of the wind: 50 mph
Explanation:
Let p = the speed of the plane
and w = the speed of the wind
It takes the plane 3 hours to go 600 miles when against the headwind and 2 hours to go 600 miles with the headwind. So we set up a system of equations.
600
m
i
3
h
r
=
p
−
w
600
m
i
2
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r
=
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+
w
Solving for the left sides we get:
200mph = p - w
300mph = p + w
Now solve for one variable in either equation. I'll solve for x in the first equation:
200mph = p - w
Add w to both sides:
p = 200mph + w
Now we can substitute the x that we found in the first equation into the second equation so we can solve for w:
300mph = (200mph + w) + w
Combine like terms:
300mph = 200mph + 2w
Subtract 200mph on both sides:
100mph = 2w
Divide by 2:
50mph = w
So the speed of the wind is 50mph.
Now plug the value we just found back in to either equation to find the speed of the plane, I'll plug it into the first equation:
200mph = p - 50mph
Add 50mph on both sides:
250mph = p
So the speed of the plane in still air is 250mph.
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