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IgorLugansk [536]
3 years ago
10

What is the estimate of 11,942 ​

Mathematics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

10000 (1sf)

12000(2sf)

11900(3sf)

11940(4sf)

Step-by-step explanation:

I would go for 12000 (2sf) if it the degree of accuracy isn't specified in the question

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Given that sin(90 -5ϴ) = Cos(180 - ϴ), Find the value of ϴ<br>​
lilavasa [31]

Answer:

30°

Step-by-step explanation:

\because \:  \sin(90 \degree -  \theta) =  \cos \theta \\ \therefore \:  \sin(90 \degree - 5 \theta) =  \cos 5\theta....(1) \\  \\ \sin(90 \degree - 5 \theta) = \cos(180 \degree -  \theta)...(2) \\ from \: (1) \: and \: (2) \\ \cos 5\theta = \cos(180 \degree -  \theta) \\ 5 \theta = 180 \degree -  \theta \\  5 \theta  + \theta= 180 \degree  \\ 6\theta= 180 \degree  \\  \theta =  \frac{180 \degree}{6}  \\  \\  \huge \red{ \boxed{\theta = 30 \degree}}

8 0
3 years ago
100 points which line is linear model​
Svetach [21]

Answer:

number 4/ the one that is in the bottom right corner. Thanks for the 100 pts.

pls rate best. God bless!

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
The cosine of the complimentary angle to 58 degrees
lana [24]
Yes
Divide it by sixty and multiply
I hope this helps
5 0
3 years ago
Which of the following equations is modeled by the graph.
Sergio [31]

Answer:

a = 50t

Step-by-step explanation:

The graph goes through (0,0) so the y intercept is 0

a=5t

a=50t

a=50 + t    not it

a=10t ​

The slope is  y/x since the y intercept is 0

m = 50/1 = 50

The equation for the line is

y = 50x+0

y = 50x

Using the variables given

a = 50t

5 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
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