All categories

3 years ago

YZ has endpoints Y(0,5) and Z (12,3). Find the length of YZ to the nearest tenth.

2 answers:

4 0

Using the formula;
$d(Y, Z) =\sqrt{(xZ - xY)^2 + (yZ - yY)^2$

You should get 12.1655 or 12.2 units.

Hope this helps you with Coordinate Solving!

larisa [96]3 years ago

3 0

Answer: The answer is 2√37 units.

Step-by-step explanation: Given that the line segment YZ has endpoints Y(0, 5) and Z(12, 3). We are to find the length of the line segment YZ.

The length of YZ will be the distance between the endpoints Y and Z.

The distance between Y and Z will be given by

$YZ=\sqrt{(0-12)^{2}+(5-3)^2}=\sqrt{144+4}=\sqrt{148}=2\sqrt{37}=12.16\sim 12.1.$

Thus, the length of YZ is 12.1 units.

You might be interested in

Find the measure of Angle b - ONLY TYPE THE NUMBER (if you write the number with anything else, it will be counted incorrect by

Alina [70]

Answer:

Step-by-step explanation:

90 degrees - 25 degrees = 65 degrees

5 0

2 years ago

Could someone help me with this problem!:)

kow [346]

Answer:

i think it's 9 just correct me if I'm wrong

4 0

2 years ago

There are 110 students in the 6th grade band and 120 students in the 7th grade band. Of the students, 60% of the 6th grade band

r-ruslan [8.4K]

Answer:

36 more students of grade 7 went on a trip than students of grade 6

Step-by-step explanation:

60 % of 110 students= 60/100*110= 66 students

85 % of 120 students= 85/100 * 120= 102 students

No of students of grade 7th more than 6th grade students= 102-66= 36

6 0

3 years ago

Please help to find an explicit formula for calculating the sum Mn

defon

The explicit formula for calculating the sum is

$S_N=\frac{n(n-1)}{2} \cdot\frac{n(n+1)}{2}$

The sum of the nth term of a sequence is expressed as;

$S_n=\frac{n}{2}(2a+(n-1)d)$

a is the first term

d is the common difference

n is the number of terms

For the sequence 0 + 1 + 2 + 3 +...

$S_n=\frac{n}{2}(2(0)+(n-1)1)\\S_n= \frac{n}{2}(n-1)\\S_n= \frac{n(n-1)}{2}$

Similarly for the sequence:

1 + 2+ 3 + 4+...

$S_n=\frac{n}{2}(2(1)+(n-1)1)\\S_n= \frac{n}{2}(2+n-1)\\S_n= \frac{n(n+1)}{2}$

Taking the product of the sum to get the explicit formula for calculating the sum

$S_N=\frac{n(n-1)}{2} \cdot\frac{n(n+1)}{2}$

Learn more here: brainly.com/question/24547297

8 0

2 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

3 years ago