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PSYCHO15rus [73]
2 years ago
6

Tan x=3/5 nearest tenth

Mathematics
1 answer:
dolphi86 [110]2 years ago
7 0

Answer:

x\approx31\°

Step-by-step explanation:

Given expression:

\tan x=\frac{3}{5}

To solve for x to nearest tenth.

Solution:

We have:

\tan x=\frac{3}{5}

In order to solve for x we will tan inverse tan on both sides in order to isolate x.

\tan^{-1}(\tan x)=\tan^{-1}\frac{3}{5}

x=\tan^{-1}\frac{3}{5}      [ Since \tan^{-1}(\tan \theta)=\theta]

x=30.96\°

We will round to nearest tenth. So,

x\approx31\°  (Answer)

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(-4,2) is a solution to the following graph. True or False?
emmasim [6.3K]
Plugging in the values of (-4,2) into the two inequality statements to see whether it is part of the solution to the inequality system:

y ≥ -5x + 3
2 ≥ -5(-4) + 3
2 ≥ 20 + 3
2 ≥ 23 (false statement).

For the other inequality statement, substitute 2 for the y value:

y > -2
2 > -2 (true statement).

Since (-4,2) only satisfies one of the inequality statements, then it means that it is not a solution to the graph.

Therefore, the correct answer is FALSE.

Please mark my answers as the Brainliest if you find my explanation helpful :)
5 0
2 years ago
Part 3: Write the equation of and graph an ellipse.
tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

The center is halfway between vertices, at (4, -6).

It is also halfway between foci.

:::::

The vertices are vertically aligned, so the parabola is vertical.

General equation for a vertical ellipse:

 (y-k)²/a² + (x-h)²/b² = 1

with

 center (h,k)

 vertices (h,k±a)

 co-vertices (h±b,k)

 foci (h,k±c), c² = a²-b²

Apply your data and solve for h, k, a, and b.

center (h,k) = (4, -6)

h = 4

k = -6

vertices (4,-6±a) = (4,-6±9)

a = 9

foci (4,-6±c) = (4,-6±5√2)

b² = a² - c² = 9² - (5√2)² = 31

b = √31

The equation becomes

 (y+6)²/81 + (x-4)²/31 = 1

:::::

length of major axis = 2a = 18

length of minor axis = 2b = 2√31

8 0
2 years ago
A grocery store chain has been tracking data on the number of shoppers that use coupons. The data shows that 71% of all shoppers
sesenic [268]

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6 0
3 years ago
Read 2 more answers
find the volume of cubical box . if the cost of painting of outer surface is rs 1440 at the rate rs 15 per mcube
bearhunter [10]
Surface of a cubical box=6(side²)

1)We have to calculate the surface of this cubical box.
Rate=cost of painting / surface  ⇒surface=cost of painting/rate

Data:
Rate=$15/m²
cost of painting=$1440

Surface=$1440/($15/m²)=96 m²

2)We find out the length of the side:

Surface of a cubical box=6(side²)

Data:
Surface of a cubical box=96 m2

Therefore:
96m²=6 (side²)
side²=96 m²/6
side²=16 m²
side=√(16 m²)=4 m

3) We find the volume of a cubical box.
volume=(side³)
volume=(4 m)³
volume=64 m³

Answer: the volume of this cubical box would be 64 m³.

3 0
3 years ago
42=3(2-3h) can someone plz answer and show me how to do this
maria [59]
You have to use distributive property.<span />
4 0
2 years ago
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