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prohojiy [21]
3 years ago
11

It took 3.5 hours for an airplane traveling at a constant speed of 550 miles per hour to reach its destination. What distance di

d the airplane travel until it reached its destination?
Mathematics
1 answer:
Nata [24]3 years ago
7 0
The formula that we want to use here is S = D/T.

S = Speed
D = Distance
T = Time

So, we know that S = 550 and T = 3.5, so we can plug them in. 

We get 550 = D/3.5 
Multiply by 3.5 on each side, and you figure out that the plane has traveled 1925 miles.
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Use your knowledge of the angle sum theorem to solve for x and find the missing angle of the triangle:
aev [14]

Answer:

x=20

Step-by-step explanation:

Well we know the right angle is 90 so...

And we also know that all the angle of a triangle add up to 180°.

Knowing this you can subtract 40° and 90°, this gives you 50°...

Then you can substitute 20° for x and then it will be 2(20°)+10°

Simplify and get 40°+10°

This equals 50° degrees

You then add that all up and you get 180°, this mean the answer is correct!

Hope this helps!

All the love, Ya boi Fraser :)

3 0
3 years ago
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What is the scientific notation of 0.005
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Move the decimal to the right 3 places: 5 x 10⁻³ 
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3 years ago
Can somebody please walked through this, I'm so confused and I have a test in 6 hours...
erastova [34]

Answer:

Q3: x = 4, y = 4, z = 4

Q4: x = 6, y = 0, z = -4

Step-by-step explanation:

Question 3: Simultaneous equations requires us to solve for x, y and z.

Since all three equations have a z in them, I will first solve for z.

Substitute in the first and third equation into the second equation.

First equation: x = 5z - 16

Second Equation: -4x + 4y - 5z = -20

Third equation: y = -z + 8

Substituting in x = 5z - 16 and y = -z + 8 for the x and y in the second equation.

-4(5z - 16) + 4(-z + 8) - 5z = -20

Expand

-20z + 64   - 4z + 32   - 5z = -20

Simplify and solve for z by putting all the numbers on one side and all the z's on the other side of the equals

-20z - 4z - 5z = -20 - 32 - 64

-29z = -116

z = -116/-29

z = 4

Substitute in this z value into the first and last equation and then solve for x and y

x = 5z - 16

x = 5(4) - 16

x = 20 - 16

x = 4

And

y = -z + 8

y = -(4) + 8

y = 4 (Its just a coincidence that they all equal to 4, I promise)

Question 5: A little bit harder of a question. Since the first and second equation both only have y and z, we can solve it using the elimination method.

Rearrange them so that the letters are on one side and numbers on the other side.

First equation: y + 6z = -24

Second equation: z + 2y = -4

I will choose to eliminate the y (You can choose either or)

Multiply the first equation by 2

2(y + 6z = -24)

2y + 12z = -48

Now that 2y is in both equations, we can minus one equation from the other to eliminate the y (I will minus the second from the first)

First Eq: 2y + 12z = -48

Second Eq: z + 2y = -4

2y - 2y = 0y

12z - z = 11z

-48 - (-4) = -44

Type these answers into a new equation

0y + 11z = - 44

Since y is 0, ignore it. Solve for z

11z = -44

z = -44/11

z = - 4

Substitute our z into either the first or second equation and solve for y (It doesnt matter which one you choose, I just did the second equation)

z + 2y = -4

(-4) + 2y = -4

2y = -4 + 4

2y = 0

y = 0

Substitute in our y and z values into the third equation and solve for x

-6x - 6y - 6z = -12

-6x - 6(0) - 6(-4) = -12

-6x - 0 + 24 = -12

-6x = -12 - 24

-6x = -36

x = -36/-6

x = 6

6 0
3 years ago
Read 2 more answers
A line contains the point ( 3, 4). If the slope of the line is -2/3write the equation of the line using slope-intercept form.
mina [271]
You use point-slope form, which is y-y1=m(x-x1) , so it becomes 
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You can make it 3y-12=-2x+6.
Add 12 to both sides to make it 3y=-2x+18.
Lastly, divide everything by 3, so it becomes y=-2/3x+6. 
7 0
3 years ago
Need answer quickly! thank you in advance!
anyanavicka [17]

Answer:

b)(b²-a²)

Step-by-step explanation:

a cotθ + b cosecθ =p

b cotθ + a cosecθ =q

Now,

p²- q²

=(a cotθ + b cosecθ)² - (b cotθ + a cosecθ)²     [a²-b²=(a+b)(a-b)]

=(acotθ+bcosecθ + bcotθ+ acosecθ) (a cotθ + bcosecθ -bcotθ-acosecθ)

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ)+b (cosecθ-cotθ)}

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} [a (cotθ-cosecθ) + {- b (cotθ-cosecθ)} ]

={a(cotθ+cosecθ)+b(cotθ+cosecθ)} {a (cotθ-cosecθ) - b (cotθ-cosecθ)}

={(cotθ+cosecθ)(a+b)} {(cotθ-cosecθ) (a-b)}

=(cotθ+cosecθ) (a+b) (cotθ-cosecθ) (a-b)

=(cotθ+cosecθ) (cotθ-cosecθ) (a+b) (a-b)        

= (cot²θ-cosec²θ) (a²-b²)                                 [(a+b) (a-b)= (a²-b²)]

= -1 . (a²-b²)                               [ 1+cot²θ=cosec²θ ; ∴cot²θ-cosec²θ=-1]

=(b²-a²)

6 0
4 years ago
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