Please helpppppppppppp

Mathematics

2 answers:

morpeh [17]2 years ago

7 0

Answer:

7.7 km

Explanation:

Use cosine rule as here given two sides and one angle.

Cosine rule states:

a² = b² + c² - 2bc cos(A)

While solving, treat a = 7.5 km as to that opposite angle is given of 68°

then b = missing side, c = 5.2 km, A = 68°

Applying rule:

7.5² = b² + 5.2² - 2(b)(5.2) cos(68)

56.25 = b² + 27.04 - 3.8959b

56.25 - 27.04 = b² - 3.8959b

b² - 3.8959b = 29.21

b² - 3.8959b - 29.21 = 0

apply quadratic equation, Here [a = 1, b = - 3.8959, c = -29.21]

$\sf b = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \quad\:when \:\ ax^2 + bx + c = 0$

$\sf b = \dfrac{ -(-3.8959) \pm \sqrt{(-3.8959)^2 - 4(1)(-29.21)}}{2(1)}$

$\sf b = 7.69 291 \quad or \quad b = -3.797$

$\sf b = 7.7 \quad (rounded \ to \ nearest \ tenth)$

As length cannot be negative. Hence the value of b is only 7.7 km

elena55 [62]2 years ago

4 0

The answer is 7.7 km.

Let's apply the Cosine Law in this situation.

Now, substitute the values based on the given diagram.

(7.5)² = b² + (5.2)² - 2(b)(5.2)(cos 68°)
56.25 = b² + 27.04 - 3.896b
b² - 3.896b - 29.21 = 0

Here, using the Quadratic Equation, we can solve :

b = 3.896 ± √(3.896)² - 4(1)(-29.21) / 2
b = 3.896 ± √15.178816 + 116.84 / 2
b = 3.896 ± √132.018816 / 2
b = 3.896 + 11.49 / 2
b = 7.7 km

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Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1