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3 years ago

Four members of a tennis team equally shared 2 granola bars at the end of practice. how much of a granola bar did they get?

Mathematics

1 answer:

Rama09 [41]3 years ago

7 0

4 divided by 2 is 2. So, 1/2 of one bar goes to each player.

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An engineer needs to determine the amount of stone to order for a road construction project. The table shows the amount of stone

Elis [28]

Answer: 1013, 2962, and 7350

Step-by-step explanation:

you move the decimal point to the right until you get to the thousands place.

8 0

2 years ago

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Help me and ill give brainlist

GrogVix [38]

its a sqaure so the angles are 90 degress each.

90 - 38 = 52

answer: C. 52

6 0

2 years ago

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gal created a painting with an area of 56 square inches and a length if 7 inches. they create a second painting with an area of

qwelly [4]

Answer:

5 inches

Step-by-step explanation:

the length of painting is 56/7 = 8in. to find the length of the 2nd painting divide the area by the width: 40/8 = 5

3 0

1 year ago

Express 194720 in standard form correct to 3 significant figures<br>

ELEN [110]

Answer:

1.95 x 10^5

Step-by-step explanation:

First look at the 4th digit in the number 194720 as the the question said to round to 3 significant figures. If the 4th digit is more than 5, then add 1 to the third digit which results in 195000. Now the question also said to express in standard form. Look for the first non-digit number in 195000. After 1 put a decimal point and remove all the zeros after the point which results in 1.95. Now count how many digits you moved to the left which comes as 5. Write the answer as 1.95 x 10^5.

Hope you can understand this!!

3 0

1 year ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

1 year ago