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3 years ago

What is the length of the hypotenuse of a right triangle with legs or 5cm and 6cm

Mathematics

2 answers:

Oksana_A [137]3 years ago

6 0

Answer:

hypotenuse = squareroot of 61 cm. which is approximately 7.81025 cm

Step-by-step explanation:

5 squared + 6 squared = hypotenuse squared

square root of (5 squared + 6 squared)= hypotenuse

zvonat [6]3 years ago

4 0

7.81

use the formula a^2+b^2=c^2
put number in formula accordingly
5^2+6^2=c^2

solve: 25+36=61

find the square root of 61

answer is 7.81

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Gnesinka [82]

Answer:

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Step-by-step explanation:

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3 years ago

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Dainel paid $22 for a water bottle. This amount includes 5% tax. What was the cost of the item befor tax?

Oksana_A [137]

20.90 would be the item before tax. To get the amount of tax, I multiplied 20 by 0.05 and subtracted that amount from 22

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3 years ago

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In ΔUVW, w = 3 inches, ∠W=23° and ∠U=73°. Find the length of u, to the nearest 10th of an inch.

kirza4 [7]

Answer:

<h2>7.4inches</h2>

Step-by-step explanation:

Check the attachment for the diagram. Sine rule will be used to get the unknown side of the triangle.

According to the rule;

$\frac{u}{sinU} = \frac{v}{sinV} = \frac{w}{sinW}\\\frac{u}{sinU} = \frac{w}{sinW}$

Given w = 3 in, ∠W=23° and ∠U=73°, on substituting into the equation above to get u we have;

$\frac{u}{sin73^{0} } = \frac{3}{sin23^{0} }\\usin23^{0} = 3sin73^{0}\\u = \frac{3sin73^{0} }{sin23^{0} }\\u = \frac{2.87}{0.39} \\u = 7.358\\u = 7.4in$

The length of u is 7.4inches to nearest 10th of an inch

6 0

3 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

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

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

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

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

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

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

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

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3 years ago

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Y=2x+1

yan [13]

Answer:

(1) 2 (2) (-1/2,0) (3) (0,1)

Step-by-step explanation:

The slope of the line is the number times x. This equation is y=mx+b, where m is the slope and b is the y-intercept. In this case, m is 2, so we have our slope. The y-intercept is easy, as we already know it to be (0,1). The x-intercept is the point where the line hits x when y=0. To solve for the x-intercept, we set y to 0 and solve. We have 0=2x+1. First, we subtract 1 from both sides and get -1=2x. Next, to get x by itself, divide both sides by 2. Now we have -1/2=x. Now we have our x coordinate for our x-intercept. Because of this, we get (-1/2,0) as our x-intercept.

3 0

3 years ago