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4 years ago

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Find the measure of each acute angle in a right angle where the measure of one acute angle is 5 times the measure of the other a

cute angle

1 answer:

lubasha [3.4K]4 years ago

7 0

We are given that the measure of one acute angle is 5 times the measure of the other acute angle.

Let us assume the measure of the other acute angle = x degree.

First acute angle is 5 times the other, that is = 5x degrees.

Both angles are making a right angle.

A right angle sum upto 90 degrees.

Therefore,

x+5x = 90.

6x = 90.

Dividing both sides by 6, we get

x =15.

<h3>Therefore, other acute angle = 15° and one angle measure = 5 times of 15 that is 75°</h3>

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Factor and solve the equation x^3-x^2-30x=0

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Answer:

Step-by-step explanation:

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3 years ago

What is an equation of the line that passes through the points (-6, -2) and

den301095 [7]

Answer:

The equation of line is: $\mathbf{4x-3y=-18}$

Step-by-step explanation:

We need to find an equation of the line that passes through the points (-6, -2) and (-3, 2)?

The equation of line in slope-intercept form is: $y=mx+b$

where m is slope and b is y-intercept.

We need to find slope and y-intercept.

Finding Slope

Slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have $x_1=-6,y_1=-2, x_2=-3, y_2=2$

Putting values and finding slope

$Slope=\frac{2-(-2)}{-3-(-6)}\\Slope=\frac{2+2}{-3+6} \\Slope=\frac{4}{3}$

So, we get slope: $m=\frac{4}{3}$

Finding y-intercept

Using point (-6,-2) and slope $m=\frac{4}{3}$ we can find y-intercept

$y=mx+b\\-2=\frac{4}{3}(-6)+b\\-2=4(-2)+b\\-2=-8+b\\b=-2+8\\b=6$

So, we get y-intercept b= 6

Equation of required line

The equation of required line having slope $m=\frac{4}{3}$ and y-intercept b = 6 is

$y=mx+b\\y=\frac{4}{3}x+6$

Now transforming in fully reduced form:

$y=\frac{4x+6*3}{3} \\y=\frac{4x+18}{3} \\3y=4x+18\\4x-3y=-18$

So, the equation of line is: $\mathbf{4x-3y=-18}$

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3 years ago

Write an expression that equals 24. Include an exponent.

kipiarov [429]

8x8x8 or 8 with the power of 3

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3 years ago

How can these models be used to find the sum 1.8 + 1.56?

enyata [817]

Answer:

The sum of 1.8 and 1.56 is 3.36 according to the model

Sum of decimals

Given the expression 1.8 + 1.56

Using the partial sum method

1.8 + 1.56 = (1.0 + 0.8) + (1.0+0.56)

Regroup as whole and decimal number

1.8 + 1.56 = (1.0 + 1.0) + (0.8 +0.56)

1.8 + 1.56 = 2.0 + 1.36

1.8 + 1.56 = 3.36

Hence the sum of 1.8 and 1.56 is 3.36 according to the model

3 0

2 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:

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

3 0

3 years ago

Read 2 more answers