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2 years ago

Plz help me either and questions 1 or 2

Mathematics

1 answer:

dexar [7]2 years ago

4 0

Multiply each side by 2.5 than add for perimeter .

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WILL GIVE BRAINLIEST

galina1969 [7]

I used substitution
The answer is a

3 0

2 years ago

Pls help dont care about the blue dot btw

alexdok [17]

Answer:

the answer is B

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

What is the midpoint of the line segment with endpoints (–1, 7) and (3, –3)?

Kipish [7]

The coordinates consist of x coordinate and y coordinate
(x₁,y₁) = (-1,7)
(x₂,y₂) = (3,-3)

To find the midpoint of x coordinate, use this following formula
x midpoint = (x₁ + x₂)/2
x midpoint = (-1 + 3) / 2
x midpoint = 2/2
x midpoint = 1

To find the midpoint of y coordinate, use this following formula
y midpoint = (y₁ + y₂)/2
y midpoint = (7 + (-3))/2
y midpoint = (7 - 3)/2
y midpoint = 4/2
y midpoint = 2

ANSWER
The midpoint is (1,2)

6 0

3 years ago

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

Rewrite using the distributive property then solve using gcf. 5(4+3)

Vitek1552 [10]

Answer:

Step-by-step explanation:

5×4+5×3

20+15

5 0

2 years ago