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Arturiano [62]
3 years ago
7

Find all the missing sides and angles of this triangle.

Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

A = 20°

AC = 6.6

BC = 2.4

Step-by-step explanation:

Given:

B = 70°

C = right angle = 90°

AB = 7

Required:

A, AC, and BC

Solution:

✔️A = 180 - (90 + 70) (sum of triangle)

A = 20°

✔️Use trigonometric function to find AC:

Refernce angle = 70°

Opp = AC

Hypotenuse = 7

Apply SOH,

sin 70 = Opp/Hyp

sin 70 = AC/7

7 * sin 70 = AC

6.57784835 = AC

AC = 6.6 (nearest tenth)

✔️Use trigonometric function to find Bc:

Refernce angle = 70°

Adj = BC

Hypotenuse = 7

Apply CAH,

cos 70 = Adj/Hyp

cos 70 = BC/7

7 * cos 70 = BC

2.394141 = BC

BC = 2.4 (nearest tenth)

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What is 18 less than the quotient of a number and 3 as an expression
makvit [3.9K]

Answer:

n/3 - 18

Step-by-step explanation:

"quotient of a number and 3" is n/3, and so

"18 less than the quotient of a number and 3" as an expression is:

   n/3 - 18

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Yolanda closed on a 20 year home loan for $83,000. she chooses to buy only 1 point at closing. by buying a point at her closing,
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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
4 years ago
Simplify by expressing fractional exponents instead of radicals
igomit [66]

Answer:

a^{\frac{1}{2}}b^{\frac{1}{2}}

Step-by-step explanation:

Given:

The expression in radical form is given as:

\sqrt{ab}

We need to express this in fractional exponent form.

We know that,

\sqrt a=a^{\frac{1}{2}}

Also, \sqrt{ab}=\sqrt a\times \sqrt b

Now, clubbing both the properties of square root function, we can rewrite the given expression as:

\sqrt{ab}=\sqrt a \times \sqrt b\\\\\sqrt{ab}=a^{\frac{1}{2}}\times b^{\frac{1}{2}}\\\\\therefore \sqrt{ab}=a^{\frac{1}{2}}b^{\frac{1}{2}}

So, the given expression in fractional exponents form is a^{\frac{1}{2}}b^{\frac{1}{2}}.

3 0
3 years ago
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