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Solnce55 [7]
4 years ago
13

Hi.i need help with this.i will apreciate the answer:)))

Mathematics
1 answer:
Anvisha [2.4K]4 years ago
7 0
Here is how to work 2, 3 , and 4.

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What is 14 1/4 - 13 5/6 =
svlad2 [7]

Answer:5/12 (Decimal: 0.416667)

Step-by-step explanation:

3 0
3 years ago
What is the parallel line and slope of a perpendicular like for 3x+5y=4
jonny [76]

Answer:

a line parallel has the same slope and different y-intercept, and the slope of the perpendicular line would be -3/5

Step-by-step explanation:

3x + 5y = 4

5y = -3x + 4

y = -3/5x + 4/5

8 0
3 years ago
A kite is stuck in a 36-ft tree. If the angle of elevation from the kite and the ground is 12, find the length of the kite.
yanalaym [24]

Answer:

173.15 ft.  

Step-by-step explanation:

Please find the attachment.

Let L be the length of kite.  

We have been given that a kite is stuck in  36-ft tree. The angle of elevation from the kite and the ground is 12.

We can see from our attachment that kite and tree form a right triangle. height of tree is opposite of angle of elevation and length of kite is hypotenuse.

Since we know that Sine relates the opposite and hypotenuse of a right triangle, so we will use sine to find the length of kite.

sin(12)=\frac{36}{L}    

L=\frac{36}{sin(12)}

L=\frac{36}{0.207911690818}  

L=173.1504364105882792\approx 173.15

Therefore, the length of kite is 173.15 ft.

4 0
3 years ago
Hi i really need help with these 2 questions please!
djyliett [7]

Answer:

Look at attachment

4 0
3 years ago
Read 2 more answers
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
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