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ololo11 [35]
3 years ago
9

Cost A=0.6489 take the inverse cosine of both sides

Mathematics
1 answer:
Stolb23 [73]3 years ago
6 0
I am not quite sure what the question asks for,
But this is what i assume it wants:

Cos A= 0.6489
In this given one, we basically find the size of the angle A
we do cosine inverse on both sides to get the size of the angle A
cos^{-1} : It looks like this in the calculator
cos^{-1} × cos A=cos^{-1}(0.6489)
(cos^{-1} and cos cancels out)
A=cos^{-1}(0.6489)
A=49.54°
check: 
cos 49.54=0.6489 (its right!)

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natita [175]
Nope .42 is greater, .4 is closer to 1 then .2 have a great day and hope this helps
4 0
3 years ago
Read 2 more answers
For what values of h and k does the linear system have infinitely many solutions? 2x1 + 5x2 = −1 hx1 + kx2 = 7
Kruka [31]

Answer:

The values of h and k for a linear system with infinitely many solutions are -2 and 5, respectively.

Step-by-step explanation:

Let 2\cdot x_{1}+5\cdot x_{2} = -1\cdot h\cdot x_{1} + k\cdot x_{2} = 7, if this linear system has infinitely many solutions, then the following conditions must be met:

-1\cdot h = 2 and k = 5.

h = -2 and k = 5

The values of h and k for a linear system with infinitely many solutions are -2 and 5, respectively.

4 0
3 years ago
Pls help, I've been stuck on this for a week
crimeas [40]

Answer:

\sin(C)=\pm\sqrt{1-\cos^2(C)}

Step-by-step explanation:

Given the Pythagorean Identity:

\sin^2(C)+\cos^2(C)=1\\

We want to solve for sin(C).

First, we will subtract cos²(C) from both sides:

\sin^2(C)=1-\cos^2(C)

Next, we will take the square root of both sides. Since we are taking an even-root, we will need to add plus/minus. Hence:

\sin(C)=\pm\sqrt{1-\cos^2(C)}

5 0
3 years ago
A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
AURORKA [14]

Answer:

For maximum area, all of the wire should be used to construct the square.

The minimum total area is obtained when length of the wire is 10m

Step-by-step explanation:

For maximum,  we use the whole length

For minimum,

supposed the x length was used for the square,

the length of the side of the square = x/4m

Area = \frac{x^{2} }{16}

For the equilateral triangle, the length of the side =  \frac{23 - x}{3}

Area = \frac{\sqrt{3} }{4}  a^{2}  = \frac{\sqrt{3} }{4} (\frac{23 - x}{3} )^{2}

Total Area = \frac{x^{2} }{16}  + \frac{\sqrt{3} }{36} (23-x)^{2}

\frac{dA}{dx}  = \frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x)\\

\frac{d^{2}A }{dx^{2} }  = \frac{1}{8}  + \frac{\sqrt{3} }{18}  > 0, therefore it is minimum

\frac{dA}{dx}  = 0 \\\\

\frac{x}{8}  -  \frac{\sqrt{3} }{18} (23 - x) = 0\\

x = 10.00m

3 0
3 years ago
Read 2 more answers
X + 5=12
slavikrds [6]
$65 if you include the $20 he already had in his account
4 0
3 years ago
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