All categories

2 years ago

I need help so can u help me ok here is the problem write a story problem with 6 divided by 3/4

Mathematics

2 answers:

Lostsunrise [7]2 years ago

5 0

0.5 is the answer for the problem

never [62]2 years ago

3 0

Here is an optional story problem:

You have a ribbon that is 6 inches long.
You need to cut sections that are each 3/4 inches long. How many sections of ribbons will you have?

The answer to this is 8 sections of ribbon.

Hope this helped!

Comment if you have any questions!!!

You might be interested in

Why was 56 and 36 points selected?

AleksAgata [21]

I don’t know , why don’t you tell me why they were selected

6 0

3 years ago

Provide one specific example of a chart and one specific example of a graph in a business situation in which you would best be a

castortr0y [4]

ONE EXAMPLE IS A GROUP OR BUSINESS MEETING SO YOU CAN DISCUSS THE DATA TOGETHER!!!!!! HOPE I ANSWERED YOUR QUESTION!

6 0

2 years ago

Solve 4(-4x- 3)> 36

wariber [46]

Answer:

x=3

Step-by-step explanation:

Simplifying

4(4x + -3) = 36

Reorder the terms:

4(-3 + 4x) = 36

(-3 * 4 + 4x * 4) = 36

(-12 + 16x) = 36

Solving

-12 + 16x = 36

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '12' to each side of the equation.

-12 + 12 + 16x = 36 + 12

Combine like terms: -12 + 12 = 0

0 + 16x = 36 + 12

16x = 36 + 12

Combine like terms: 36 + 12 = 48

16x = 48

Divide each side by '16'.

x = 3

Simplifying

x = 3

8 0

3 years ago

Read 2 more answers

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

How do you solve this equation

Eva8 [605]

X= 1/2 or x = 3/2.
Step 1: factor left side of equation. (2x-1)(2x-3)=O
Step 2: Set factors equal to 0. 2x-1=0 or 2x-3=0
X= 1/2 or x = 3/2.

5 0

3 years ago