All categories

3 years ago

How can you check the answer to a decimal multiplication problem

Mathematics

2 answers:

frez [133]3 years ago

4 0

The same way you would if there was no decimal

Svet_ta [14]3 years ago

4 0

1.Multiply normally, ignoring the decimal points.
2.Then put the decimal point in the answer - it will have as many decimal places as the two original numbers combined.

You might be interested in

Simplifique as expressões 2-1+2-2+36-1/2 (81)3/4+(16)-1/4-1-100

Dovator [93]

$- \frac{629}{8} or -78.625 or -78 \frac{5}{8}$

4 0

3 years ago

The velocity of a particle moving along a line is given by v(t) = e ^ t * sin(e ^ t) the interval 0<= t<= pi 2 . What is t

Butoxors [25]

Answer:

0.44237 units

Step-by-step explanation:

Given

$v(t) = e^t*sin(e^t)$

$0 \le t \le \frac{\pi}{2}$

Required

The distance traveled in this interval

We have:

$v(t) = e^t*sin(e^t)$

The distance is calculated as:

$d(t) = \int\limits^a_b {v(t)} \, dt$

So, we have:

$d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt$

Let:

$u = e^t$

Differentiate

$\frac{du}{dt} = e^t$

So:

$dt = e^{-t} du$

So, we have:

$d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt$

$d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(u)} \, e^{-t} du$

Rewrite as:

$d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t* e^{-t}*sin(u)} \, du$

$d(t) = \int\limits^{\frac{\pi}{2}}_0 {sin(u)} \, du$

Integrate:

$d(t) = -\cos(u)|\limits^{\frac{\pi}{2}}_0$

Substitute $u = e^t$

$d(t) = -\cos(e^t)|\limits^{\frac{\pi}{2}}_0$

Split

$d(t) = -\cos(e^\frac{\pi}{2}) - [-\cos(e^0)]$

$d(t) = -\cos(e^\frac{\pi}{2}) +\cos(e^0)$

$d(t) = -0.09793 + 0.5403$

$d(t) = 0.44237$

The distance traveled in this interval is: 0.44237 units

dt =

d(t) = \int\limits^2_0 {e^t*sin(e^t)} \, dt

4 0

3 years ago

How to solve ,step by step

Serggg [28]

Answer:

gradient = slope = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ = $\frac{rise}{run}$

Slope intercept Form equation : y = mx + b

m = slope or gradient

b = y - intercept ( where the line crosses the y = axis)

x and y = are place holders for a coordinate pair that makes the equation true

c) y = -6x + 8

The -6 is the m. It's the slope or gradient.

the + 8 is the b. It's the y- intercept.

d) y = 4

This is a horizontal line. It intercepts the y-axis at 4.

That means the 4 is the y-intercept.

There is no x. That means the slope is 0. The line rises 0 as it runs left to right.

e) y -4x= 0

equation needs to be is standard form y = mx + b.

add 4x to both sides in order to isolate the y variable.

y = 4x + 0.

The slope or gradient is 4. The y - intercept is 0. The line crosses through the origin.

f) y -x = -8

Add x to both sides.

y = x - 8

There is one x. That means the gradient is 1. The y-intercept is the -8

g) y + 3x = 7

Subtract 3x from both sides.

y = -3x + 7

-3 = gradient. 7 = y-intercept.

h) y + $\frac{1}{2}$ x = -4

Subtract $\frac{1}{2}$ x from each side.

y = - $\frac{1}{2}$ x - 4

One last thing. If you are presented with an equation without a y, the gradient is 'undefined'.

example : x = 4

This a vertical line passing through 4 on the x-axis. There is no 'b' because its not crossing the y-axis.

Why is it 'undefined' ?

As the line rises it, it does not 'run' left or right. $\frac{rise}{0}$ . Zero can never, ever be in the denominator. Denominators can't be zero. That is why we say it's 'undefined'.