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3 years ago

What the awnser 2/4 x 4/8

Mathematics

2 answers:

omeli [17]3 years ago

6 0

Answer: 1/4 or 0.25.

Step-by-step explanation: 2/4 x 4/8 = 8/32 = 1/4 = 0.25.

aleksandrvk [35]3 years ago

3 0

Answer:

1/4 or 0.25

Step-by-step explanation:

2/4 * 4/8 =

2*4=8 and 4*8=32

so 2/4 * 4/8 = 8/32

8/32 = 1/4 = 0.25

So the answer is:

1/4 or 0.25

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Please tell me if this is right or wrong I really need to know.

katrin2010 [14]

Answer:

8 ft

Step-by-step explanation:

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3 years ago

The cost of taking your pet aboard the air flight with you in the continental US varies according to the airlines. The five numb

sergeinik [125]

Answer:

Lowest is 100

Highest is 125

Step-by-step explanation:

We use the 5 number summary to be the foundation of a graphical representation referred to as the box plot. One box would move from one quartile which is the lowest quartile Q1 to the another quartile Q3 which is the upper quartile.

Now if a box plot is to be made given the the information in this question, the box is going to go from Quartile 1 to Quartile 3.

Then the Lowest value would be 100 and the highest 125

3 0

3 years ago

In a class of 28 pupils, if 12 are females. Calculate the ratio of males : total

tensa zangetsu [6.8K]

Answer:

16 : 28

Step-by-step explanation:

If 12 out of the 28 pupils are female, we can calculate the number of males by subtracting the number of females from the total number of pupils.

28 - 12 = 16

So, there are 16 males in the class. Now, we know that the ratio of males : total is 16 : 28.

I hope this helps! Have a lovely day!! :)

(brainliest is much appreciated!!)

6 0

2 years ago

Read 2 more answers

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

Statement: If lines are skew, then they are not coplanar.

ArbitrLikvidat [17]

The best and most correct answer among the choices provided by the question is the third choice. The contrapositive statement is " If lines are not coplanar, then they are skew." I hope my answer has come to your help. God bless and have a nice day ahead!

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4 years ago