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stich3 [128]
3 years ago
8

Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 centimeters, the length of C B is 9.8

centimeters, and the length of A C is 6.9 centimeters.
Which expressions can be used to find m∠BAC? Select three options.

cos−1(StartFraction 6.9 Over 12 EndFraction)
cos−1(StartFraction 9.8 Over 12 EndFraction)
sin−1(StartFraction 6.9 Over 12 EndFraction)
sin−1(StartFraction 9.8 Over 12 EndFraction)
tan−1(StartFraction 6.9 Over 9.8 EndFraction)

Mathematics
2 answers:
Dafna11 [192]3 years ago
3 0

Step-by-step explanation:

yo digo que es eso la verdad no se

Tcecarenko [31]3 years ago
3 0

Answer:

pls mark brainliest

Step-by-step explanation

You might be interested in
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
3 years ago
How long is each side of a square that has an area of 25 meters?
Sergio [31]
The square root of 25 is 5, so each side is 5metres x
4 0
3 years ago
Read 2 more answers
Simplify:9-1÷2÷27^2÷3​
sammy [17]

<u>First let's calculate the exponents.</u>

27^2 : 729 = 9-1\div \:2\div \:729\div \:3

<u>Now we should multiply and divide.</u>

<u />1\div2\div729\div3:\frac{1}{4374} = 9-\frac{1}{4374}<u />

<u>Now we should add and subtract.</u>

<u />9-\frac{1}{4374} :\frac{39365}{4374} =  \frac{39365}{4374}<u />

<u>Convert the improper fractions into mixed numbers.</u>

<u />8\frac{4373}{4374}<u />

<u />

<u>Answer : </u>

<u />8\frac{4373}{4374}<u />

<u />

5 0
3 years ago
The fourth term of an Arithmetic Sequence is equal to 3 times the first term, and the seventh term exceeds twice the third term
11Alexandr11 [23.1K]

Answer:

The first term is 3. The common difference is 2.

Step-by-step explanation:

The first term is x.

The common difference is d.

The second term is x + d.

3rd term: x + 2d

4th term: x + 3d

7th term: x + 6d

"The fourth term of an Arithmetic Sequence is equal to 3 times the first term"

x + 3d = 3 * x       Eq. 1

"the seventh term exceeds twice the third term by 1"

x + 6d = 2(x + 2d) + 1       Eq. 2

Simplify Eq. 1:

2x = 3d

Simplify Eq. 2:

x + 6d = 2x + 4d + 1

x = 2d - 1

Multiply both sides of the last equation by 2.

2x = 4d - 2

2x = 3d   (simplified Eq. 1)

Since 2x = 2x, then the right sides are equal.

3d = 4d - 2

d = 2

2x = 3d

2x = 3(2)

2x = 6

x = 3

Answer: The first term is 3. The common difference is 2.

8 0
3 years ago
Elsie loves to garden the width of her current garden is half the length. Elsie wants to increase both the width and the length
Pani-rosa [81]

Answer:

length of new garden: 3x or 3y times 2

width of new garden: 3y or 3x/2

Area: 3x times 3y

Step-by-step explanation:

x= original length of garden

y= original width of garden

if she triples it she needs to multiply the old measurements by 3

The area of  her new garden is the new measurements multiplied by one another

4 0
3 years ago
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