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Vera_Pavlovna [14]
3 years ago
15

Quick.!! Il pick the brainliest.!! Tha value of x is.?

Mathematics
2 answers:
aleksklad [387]3 years ago
8 0
If you add up all the angles in a triangle it will always be 180.
180 - 45x is angle TRS
Therefore 180 - 45x + 25x + 57 + x = 180
Subtract 180 from both sides
-45x + 25x + 57 + x = 0
Subtract 57 frok both sides
-45x + 25x + x = -57
Combine like terms
-19x = -57
divide both sides by -19 and you get
x = 3
Zarrin [17]3 years ago
5 0

The sum of the angles in a triangle is 180°. It means that when you add up the three angles, you would get 180°.

The angle on a straight line, is also 180°.

Since ∠QRT = 45x, then...

∠TRS = 180° - 45x

So now you add up the angles in the triangle:

(180° - 45x) + 25x + (57 + x) = 180°

Combine like terms:

-19x + 237° = 180°

Subtract 237° to both sides:

-19x = -57

Now, divide both sides by -19:

x = 3

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Step-by-step explanation:

find the change in x so +4

and the change in they y so 3

3/4 and divide that

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An item is regularly priced at $85. Tom bought it at a discount of 15% off the regular price
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$72.25

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85 x (1-.15) = answer

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Answer:

456 / 10 = <u>45.6</u>

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When two six-sided dice are rolled what is the probability that the product of their scores will be greater than six?
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Answer:   \bold{\dfrac{11}{18}}

<u>Step-by-step explanation:</u>

Think of the products row by row:

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21 22 23 24 25 26  - 3 products greater than 6

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2 years ago
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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 <em>0.002652</em> 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

<em>f(x)=cos(x) </em>

<em> </em>

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
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