All categories

3 years ago

How do you solve 5(1.24)

Mathematics

2 answers:

zzz [600]3 years ago

8 0

The answer is 6.2 your welcome :)

mezya [45]3 years ago

4 0

6.2 is the correct answer. First you had to remove parenthesis first, and it gave us, $5*1.24$ . And finally, you had to multiply by the numbers, and it gave us the answer is $5*1.24=6.2$ , and 6.2 is the final answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

You might be interested in

These triangles are not drawn to scale. Use the Pythagorean Theorem to determine which triangle is a right triangle. Question 4

Svetach [21]

Answer:

it's probably D lol BUTTTT I have no idea

8 0

3 years ago

Read 2 more answers

The endpoints of a side of rectangle ABCD in the coordinate plane are at A (3,9) and

Montano1993 [528]

Answer:

y = - 4x + 38

Step-by-step explanation:

A(3, 9)

B(5, 1)

$m_{AB}$ = $\frac{1-9}{5-3}$ = - 4

CD║AB

C(9, 2)

y - 2 = - 4(x - 9)

y = - 4x + 38

5 0

2 years ago

Given a square with two vertices of one side located at (-5, -3) and (-5, 12), in square units what is its area?

Ray Of Light [21]

The side length is 12 - -3 = 15 so the area is 15 squared = 225.

4 0

3 years ago

Look at this graph:

Andrew [12]

Answer:

Step-by-step explanation:

slope=rise/run=2/1=2

6 0

2 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