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MissTica
3 years ago
8

Please answer it in two minutes

Mathematics
1 answer:
kari74 [83]3 years ago
6 0
54 miles only due to the angles being congruent to each other if you look closely
You might be interested in
(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
Please answer correctly !!!!!! Will mark brainliest !!!!!!!!!!!
Iteru [2.4K]

Answer:

1 A

2 A

Step-by-step explanation:

1. steps can be positive whole numbers only

2. there is 15 steps

7 0
3 years ago
Find the measure of arc TV
Korolek [52]
90 degrees
Hope this helps
6 0
3 years ago
Use natural logarithms to solve the equation. Round to the nearest thousandth. 3e^2x +5=27
puteri [66]

Answer:

x=0.996

Step-by-step explanation:

3e^{2x} +5=27

To take natural log ln , we need to get e^2x alone

Subtract 5 on both sides

3e^{2x}=22

Now we divide both sides by 3

e^{2x}=\frac{22}{3}

Now we take 'ln' on both sides

ln(e^{2x})=ln(\frac{22}{3})

As per log property we can move exponent 2x before ln

(2x)ln(e)=ln(\frac{22}{3})

The value of ln(e) = 1

2x=ln(\frac{22}{3})

Divide both sides by 2

x=\frac{ln(\frac{22}{3})}{2}

x= 0.996215082

Round to nearest thousandth

x=0.996

6 0
3 years ago
Find the domain and the range of f(x)=2-x+sin x
stellarik [79]
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.<span><span>(<span><span>−∞</span>,∞</span>)</span><span><span>-∞</span>,∞</span></span><span><span>{<span>x|x∈R</span>}</span><span>x|x∈ℝ</span></span>Find the magnitude of the trig term <span><span>sin<span>(x)</span></span><span>sinx</span></span> by taking the absolute value of the coefficient.<span>11</span>The lower bound of the range for sine is found by substituting the negative magnitude of the coefficient into the equation.<span><span>y=<span>−1</span></span><span>y=<span>-1</span></span></span>The upper bound of the range for sine is found by substituting the positive magnitude of the coefficient into the equation.<span><span>y=1</span><span>y=1</span></span>The range is <span><span><span>−1</span>≤y≤1</span><span><span>-1</span>≤y≤1</span></span>.<span><span>[<span><span>−1</span>,1</span>]</span><span><span>-1</span>,1</span></span><span><span>{<span>y|<span>−1</span>≤y≤1</span>}</span><span>y|<span>-1</span>≤y≤1</span></span>Determine the domain and range.Domain: <span><span><span>(<span><span>−∞</span>,∞</span>)</span>,<span>{<span>x|x∈R</span>}</span></span><span><span><span>-∞</span>,∞</span>,<span>x|x∈ℝ</span></span></span>Range: <span><span>[<span><span>−1</span>,1</span>]</span>,<span>{<span>y|<span>−1</span>≤y≤1</span><span>}

</span></span></span>
8 0
3 years ago
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