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Elena L [17]
3 years ago
11

5+8(3+x) you have to simplify the expression. Please show your work and Explain

Mathematics
1 answer:
dedylja [7]3 years ago
6 0
Solve 5+8 and get 13(3+x) then muliply 13×3 and 13×X... you will end up with 39+13x
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Brandon sights a helicopter above a building that is 200 feet away at an angle of elevation of 30 degrees. To the nearest foot,
Firlakuza [10]
The problem says that <span>Brandon sights a helicopter above a building that is 200 feet away at an angle of elevation of 30 degrees. So, you can calculate the height asked, by following this procedure:
</span>
 Tan(α)=Opposite leg/Adjacent leg

 α=30°
 Opposite leg=x
 Adjacent leg=200 feet

 When you substitute these values into the formula above (Tan(α)=Opposite leg/Adjacent leg), you have:

 Tan(α)=Opposite leg/Adjacent leg
 Tan(30°)=x/200
 
 You must clear "x":

 x=200xTan(30°)

 Therefore, the value of "x" is:

 x=115 feet
<span>
 How high above the ground the is the helicopter?

 The answer is: 115 feet</span>
4 0
3 years ago
Using distributive property combine 13c - 7c= Show work
emmasim [6.3K]
Actually, distributive property isn't needed to combine this. You can simply subtract 7 from 13 which will give you 6. The solution is 6c.
3 0
3 years ago
Read 2 more answers
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
Is (-2,4) a solution of the graphed inequality?
k0ka [10]

we need the equation or graph to figure that out

6 0
3 years ago
Read 2 more answers
in the given diagram, which of the following can you use to prove that triangle ABC is an isosceles triangle?
ryzh [129]
If the triangle has two congruent sides.
5 0
3 years ago
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