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3 years ago

Answer please!!!!!!!!!

Mathematics

1 answer:

Sonbull [250]3 years ago

5 0

Answer:

Third term is 67.

Step-by-step explanation:

Let the first term of the sequence be $x$ .

Given:

The second term is 5 more than the first term.

Second term = $x+5$

Third term = First term + Second term = $x+x+5=2x+5$

Fourth term = Second term + Third term = $x+5+2x+5=3x+10$

Sum of first 4 terms is 237. Therefore,

$x+(x+5)+(2x+5)+(3x+10)=237\\(x+x+2x+3x) + (5+5+10)=237.........(\textrm{Commutative property of addition})\\7x+20=237\\7x=237-20\\7x=217\\x=\frac{217}{7}=31$

Now, the third term of the sequence is given as:

$\textrm{Third term}=2x+5=2(31)+5=62+5=67$

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ziro4ka [17]

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4 years ago

May I please receive help?

GrogVix [38]

Answer:

24°

Step-by-step explanation:

(4y-4)+(180-52)+3y = 180°

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3 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLEST FOR CORRECT ANSWERS. PLZ HELP ME

taurus [48]

Answer:

D slope in relationship is 3

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Step-by-step explanation:

5 0

4 years ago

A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of sixsix dig

vazorg [7]

Answer:

450,000 codes

Step-by-step explanation:

Number of digits in the access code = 6

Conditions:

First digit cannot be 2
Last digit must be odd

Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.

Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.

The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.

According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.

Therefore,

Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes

Hence, 450,000 different codes are possible for the lock box

8 0

4 years ago

The point (2, 1) is a solution to which of the following systems of equations?

Snezhnost [94]

$\huge\boxed{\left \{ {{x+y=3} \atop {3x-y=5}} \right.}$

We can solve this problem by testing the solution against each system of equations. We'll do this by substituting the point into each equation for each system.

<h2>System 1 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}2x+y&=5\\2(2)+1&=5\\4+1&=5\\5&=5\\&\checkmark\end{aligned}&\begin{aligned}-2x+y&=2\\-2(2)+1&=2\\-4+1&=2\\-3&=2\\&\text{X}\end{aligned}\end{array}$

<h2>System 2 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-2x-y&=-5\\-2(2)-1&=-5\\-4-1&=-5\\-5&=-5\\&\checkmark\end{aligned}&\begin{aligned}2x-y&=2\\2(2)-1&=2\\4-1&=2\\3&=2\\&\text{X}\end{aligned}\end{array}$

<h2>System 3 - $\Large\textbf{X}$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-x+y&=3\\-2+1&=3\\-1&=3\\&\text{X}\end{aligned}&\begin{aligned}-3x+y&=-5\\-3(2)+1&=-5\\-6+1&=-5\\-5&=-5\\&\checkmark\end{aligned}\end{array}$

<h2>System 4 - $\Large\checkmark$ </h2>

$\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}x+y&=3\\2+1&=3\\3&=3\\&\checkmark\end{aligned}&\begin{aligned}3x-y&=5\\3(2)-1&=5\\6-1&=5\\5&=5\\&\checkmark\end{aligned}\end{array}$

Since both equations in the final system are true when solved with $(2, 1)$ , the answer is the last system.

3 0

3 years ago