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2 years ago

RIGHT ANSWER I'LL GIVE BRAINLIES

Mathematics

2 answers:

Alex_Xolod [135]2 years ago

7 0

I really hope this is right but 6?

Nataly [62]2 years ago

3 0

Answer:

134

Step-by-step explanation:

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13- (8x - 5) – 16x

KatRina [158]

Answer:

The answer is 6(3-4x) because I just did on a test

3 0

3 years ago

An airport offers 2 shuttles that run on different schedules. If both shuttles leave the airport at 4.00 pm at what time will th

Ksivusya [100]

Answer:

4:18 pm

Step-by-step explanation:

An airport offers 2 shuttles that run on different schedules.If both shuttles leave the airport at 4:00 p.m.,at what time will they next leave theairport together.

Shuttle A leaves every 6 minutes

Shuttle B leaves every 9 minutes

Solution

To solve for when next both shuttles will leave the airport together, find the least common multiples of 6 minutes and 9 minutes

Shuttle A (every 6 minutes) = 6, 12, 18, 24

Shuttle B (every 9 minutes) = 9, 18, 27, 36

The least common multiples of 6 minutes and 9 minutes is 18 minutes

4:00 pm + 18 minutes

= 4:18 pm

The next time both shuttles will leave theairport together is 4:18 pm of the same day

8 0

3 years ago

Point x is located at (-2,-6) and point z is located at (0,5). Find the y value for the point y that is located 1/5 the disteanc

Sedaia [141]

Answer:

-3 1/2

Step-by-step explanation:

We have to find the distance between x and z along the y-axis.

This will be: 5 - -6 = 11 units

1/5 of 11 = 1/5 × 11

= 11/5

= 2 1/5

Now add 2.5 to -6

2 1/2 + -6 = -3 1/2

The value of y = -3 1/2

8 0

3 years ago

The equation of the line is 3y+2x=12 . what is the slope of the line perpendicular to this equation?

Ket [755]

The slope
$\frac{3}{2}$

6 0

3 years ago

Read 2 more answers

Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

6 0

1 year ago