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3 years ago

Please answer this correctly without making mistakes my work is due today

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

24.24

Step-by-step explanation:

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Given the triangle below, answer the following:

OLEGan [10]

Answer:

Step-by-step explanation:

question no . 5

ratio of angle L

take angle as reference angle

using cos rule

cos L = base / hypotenuse

= 4/5

ratio of angle N

take angle N as reference angle

using sin rule

sin N = opposite / hypotenuse

= 4/5

for measure of angle L

cos L = 4/5

cos L = 0.8

L = $cos^{-1}(0.8)$

L = 36.9

for meanure of angle N

sin N = 4/5

sin N = 0.8

N = $sin^{-1}(0.8)$

N = 53.1

7 0

3 years ago

Read 2 more answers

Find the specified element in the inverse of the given matrix

JulijaS [17]

What is the given matrix?
you didn't show a matrix.
the components of the matrix are the elements.
and the elements of the inverse matrix are the elements of the inverse matrix
1)
so if matrix A is 1 4 1
2 5 2
3 6 3
2)
14114
25225
36336

3)
15 24 12
15 12 24
4)
[A]=(15+24+12=51)-(15+12+24=51)
[A]=0
5)
  52 23 25
63 23 36

41 11 14
63 33 36

41 12 14
52 12 25
6)

15-12 6-6 12-15

12-6 3-3 6-12

8-5   2-2 5-8

7)
3 0   -3

6 0 -6

3 0 -3
8)
(+)3 (-)0 (+)-3

(-)6 (+)0 (-)-6

(+)3 (-)0 (+)-3
=

  3 0 -3
-6 0   6
3 0 -3

9)
  3 -6 3
N= 0 0 0
-3 6 -3
10) the inverse of [A]= 1/|A|*[N]
which is in this case ⁻A= 0
because |A|=0

6 0

3 years ago

What is 4 of one tenth

schepotkina [342]

0.4
4×1/10=0.4
multiply 4 by one-tenth

3 0

3 years ago

Graph ABCD if A(1, 4), B(6, 6), C(4, 1), and D(−1, −1). Is ABCD a rhombus? Show how you know.

stich3 [128]

C4,1 because I solved it and it showed me

6 0

3 years ago

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

3 0

3 years ago