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Artyom0805 [142]
3 years ago
9

How are you today? don't forget that you are loved!

Mathematics
2 answers:
Shalnov [3]3 years ago
6 0

I'M SOOOOOOOO GOOD

HOW ABOUT YOU

aliina [53]3 years ago
4 0

Answer:

I´m good, hbu?

You might be interested in
(England,
murzikaleks [220]

Answer:

∠JKL = 38°

Step-by-step explanation:

PQRS, JQK and LRK are straight lines

Let's take the straight lines in the diagrams one after the other to find what they consist.

The related diagram can be found at brainly (question ID: 18713345)

Find attached the diagram used for solving the question.

For straight line PQRS,

2x°+y°+x°+2y° = 180°

(Sum of angles on a Straight line = 180°)

Collect like terms

3x° + 3y° = 180°

Also straight line PQRS = straight line PQR + straight line SRQ

For straight line PQR,

2y + x + ∠RQM = 180° ....equation 1

For straight line SRQ,

2x + y + ∠MRQ = 180° ....equation 2

Straight line PQRS = addition of equation 1 and 2

By collecting like times

3x +3y + ∠RQM + ∠MRQ = 360°....equation 3

Given ∠QMR = 33°

∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)

∠RQM + ∠MRQ + 33° = 180°

∠RQM + ∠MRQ = 180-33

∠RQM + ∠MRQ = 147° ...equation 4

Insert equation 4 in 3

3x° +3y° + 147° = 360°

3x +3y = 360 - 147

3x +3y = 213

3(x+y) = 3(71)

x+y = 71°

∠JQP = ∠RQK = 2y° (vertical angles are equal)

∠LRS = ∠QRK = 2x° (vertical angles are equal)

∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)

2x+2y + ∠QKR = 180

2(x+y) + ∠QKR = 180

2(71) + ∠QKR = 180

142 + ∠QKR = 180

∠QKR = 180 - 142

∠QKR = 38°

∠JKL = ∠QKR = 38°

5 0
3 years ago
What is the midpoint between z1 = -8 + 3i and z2 = 4 + 7i?
I am Lyosha [343]

Answer:

Therefore, Mid point : -2+5i.

Step-by-step explanation:

Given  :  Z1 = -8 + 3i and Z2 = 4 + 7i.

To find : What is the midpoint.

Solution : We have given  Z1 = -8 + 3i and Z2 = 4 + 7i.

Mid point :  \frac{z_{1}+z_{2}}{2}.

On plugging the values.

Mid point :  \frac{-8+3i + 4 +7i}{2}.

Combine like terms.

Mid point :  \frac{-8+4+3i +7i}{2}.

Mid point :  \frac{-4+10i}{2}.

Mid point : -2+5i.

Therefore, Mid point : -2+5i.

8 0
3 years ago
If m^1=m^2 then m^ 1 is?
e-lub [12.9K]

Answer:

90

Step-by-step explanation:

5 0
3 years ago
The student council at Spendtoomuch High School is
Ivan
Answer: the won’t profit. they are in the hole $-50

explanation: if their budget is $300 and their total for the Dj and decorations is $350. they overspent money so they won’t profit. they will just be spending more than planned.
3 0
3 years ago
Help me please ! ASAP . . . .
salantis [7]

\textbf{Answer:}

\medskip\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)tan^{-1}(x)\mid_{x=0}^{x=\infty}}dy=\frac{\pi^2}{4}}

\medskip\newline\textbf{Step-by-step explanation:}

\medskip\newline\text{It's been a very long time since I've seen a calculus question.}\newline{\text{It might be because I haven't visited this site in a while.}}\text{Anyway, on to the answer.}

\medskip\newline\text{When doing 2 integrals, work inside to out and treat the variables}\newline\text{ that aren't the ones being integrated as constants and proceed like normal.}\medskip\medskip\newline\int\limits_{0}^{\infty}{\int\limits_{0}^{\infty}{\frac{1}{(x^2+1)(y^2+1)}}dx}dy

\newline{\text{Work inner to outer. Therefore, treat all y's as constants for now.}}\newline{\text{By the integration formula: }\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+C}

\newline{\int\limits_{0}^{\infty}{\int\limits_{0}^{\infty}{\frac{1}{(x^2+1)(y^2+1)}}dx}dy=}

\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)tan^{-1}(x)\mid\limits_{x=0}^{x=\infty}}dy=}

\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(tan^{-1}(\infty)-tan^{-1}(0))}dy=}

\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2}-0)}dy=}

\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2})}dy}

\medskip\newline\text{Using the same integration formula as before.}

\medskip\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2})}dy=}

\newline{(\frac{\pi}{2})tan^{-1}(y)\mid\limits_{y=0}^{y=\infty}=}

\newline{(\frac{\pi}{2})(tan^{-1}(\infty)-tan^{-1}(0))=}

\newline{(\frac{\pi}{2})(\frac{\pi}{2}-0)=}

\newline{(\frac{\pi}{2})(\frac{\pi}{2})=}

\newline{\frac{\pi^2}{4}}

\bigskip\newline{\text{Therefore, the answer is }\frac{\pi^2}{4}}

4 0
3 years ago
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