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Alinara [238K]
3 years ago
10

Find X value.Will rate brainliest for the right ans​

Mathematics
1 answer:
mina [271]3 years ago
7 0

Answer:

\huge\boxed{x=65}

Step-by-step explanation:

Look at the picture.

m\angle VUP=180^o-2\cdot60^o=180^o-120^o=\boxed{60^o}\\\\m\angle UPQ=360^o-2\cdot60^o=360^o-120^o=\boxed{240^o}\\\\m\angle QRM=180^o-2\cdot60^o=180^o-120^o=60^o\\\\m\angle RMQ=180^o-135^o=45^o\\\\in\ \triangle QRM:\ m\angle MQR=180^o-(60^o+45^o)=180^o-105^o=75^o\\\\m\angle PQW=360^o-(2\cdot60^o+75^o+80^o)=360^o-(120^o+155^o)\\\\=360^o-275^o=\boxed{85^o}\\\\m\angle UVW=\boxed{90^o}

VWQPU is the pentagon. The sum of interior angles in a pentagon is equal 540°. Therefore:

x^o=540^o-(90^o+60^o+240^o+85^o)=540^o-475^o=65^o

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Identify the domain and range of each function. Make sure to provide these answers using inequalities.
yulyashka [42]

Answer:

a. Domain: (-∞, ∞)

   Range: (0,∞)

b. Domain: (-∞, ∞)

   Range: (0,∞)

c. Domain: (-∞, ∞)

   Range: (-∞,0)

d. Domain: (-∞, ∞)

   Range: (-∞,0)

e. Domain: (-∞, ∞)

   Range: (0,∞)

Step-by-step explanation:

a. y= 3(2)^x \\b. y= 7(0.4)^x \\c. y = -2(0.6)^x \\d. y = -3(4)^x \\e. y = 2(22)^x

These equations are all exponential functions. Exponential functions are curves which approach a horizontal asymptote usually at y=0 or the x-axis unless a value has been added to it. If it has, the curve shifts. None of these have that and their y - values remain between 0 and ∞. This is the range, the set of y values.

However, the range of exponentials can change based on the leading coefficient. If it is negative the graph flips upside down and its range goes to -∞. C and D have this. Their range is (-∞, 0)

In exponential functions, the x values are usually not affected and all are included in the function. Their domain is (-∞, ∞). All of these equations have this domain.

a. Domain: (-∞, ∞)

   Range: (0,∞)

b. Domain: (-∞, ∞)

   Range: (0,∞)

c. Domain: (-∞, ∞)

   Range: (-∞,0)

d. Domain: (-∞, ∞)

   Range: (-∞,0)

e. Domain: (-∞, ∞)

   Range: (0,∞)


5 0
3 years ago
9/4 = ? divided by 4
Luda [366]

Answer:

9

Step-by-step explanation:

5 0
3 years ago
What about this one ?
salantis [7]
The options are given in factored form.

If a function is zero at x=3, this means x=3 is a root of the equation. x = 3 can also be written as x - 3 = 0

So, if x = 3 is a root of the equation, x - 3 will be a factor of the function. From the given options we can see only Option C contains x - 3 as its factor. Substitute x = 3 in option C and the function value will be zero.

So, the correct answer is option C
6 0
4 years ago
Read 2 more answers
If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams
Juliette [100K]

Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

N=N_0e^{-\lambda t}

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34, t= 10

N=N_0e^{-\lambda t}

\Rightarrow \frac {N}{N_0}=e^{-\lambda t}

\Rightarrow \frac34 =e^{-\lambda .10}

Taking ln both sides

\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|

\Rightarrow ln|\frac34|=-10\lambda

\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}

Now , N₀= 512 grams, N= 121.5 grams, t=?

N=N_0e^{-\lambda t}

\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}

\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}

\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}

Taking ln both sides

\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|

\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}

\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}

\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

8 0
4 years ago
Help me with this question please
Finger [1]

a) 1,04

b) 160*0,94=150,4

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

P.S. Hello from Russia

8 0
3 years ago
Read 2 more answers
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