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3 years ago

Find the value of the variable ??

Mathematics

2 answers:

Leviafan [203]3 years ago

7 0

Answer:

Step-by-step explanation:

eduard3 years ago

6 0

Answer:

Explanation:

The triangle is a special right triangle.

Since it has a 45° in one of the two unknown angles, the other angle is also 45°. This is called a 45-45-90 right triangle.

The rule for that special triangle is that the legs measure the same while the hypotenuse measures a multiplication of √2.

Since one leg is labeled 7 units, the other (which is x) is labeled 7 too.

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Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago

-1.2 29 20 Enter the answer as an exact decimal or simplified fraction

Rudiy27

Answer:

ohh

Step-by-step explanation:

6 0

3 years ago

How do you slove this equation 12 n3 ------- 3(n4)2

sergejj [24]

Answer:

36n

----------------

24n

8 0

3 years ago

Read 2 more answers

Please help!!!! I'll make you Brainliest!! Thanks!!

alexandr1967 [171]

Im thinking it whould be c

6 0

4 years ago

What is the domain in interval notation?

Free_Kalibri [48]

Hello!

Interval notation is written using brackets or parentheses. Brackets mean that the value of the corresponding side is equal to the function. Parentheses are the opposite of brackets in this case, and it is not part of the equation.

I will post the graph of this equation below.

Therefore, the domain of this equation in interval notation is (-∞, 6/7).

Note that it is parentheses not brackets. The graph does not go past 6/7, and infinity is not a number, it is a concept.

3 0

4 years ago