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IceJOKER [234]
3 years ago
7

Llamando x a un numero, expresa en lenguaje algebraico su doble

Mathematics
1 answer:
Rasek [7]3 years ago
3 0
In English plsYour on t
You might be interested in
Solve by substitution
storchak [24]
4x - 2y = 14
y = 1/2x - 1

4x - 2(1/2x - 1) = 14
4x - x +2 = 14
3x + 2 - 2 = 14 - 2
3x = 12
3x / 3 = 12/3
x = 4


2x = y -10
2x + 7 = 2y

y - 10 + 7 = 2y
y -y -10 + 7 = 2y - y
-3 = y


8x - 1/3 y = 0
12x + 3 = y

8x - 1/3y = 0
8x -1/3( 12x + 3) = 0
8x - 4x - 3 = 0
4x -3 + 3 = 0 + 3
4x / 4 = 3/4
x = 3/4
3 0
3 years ago
I am stuck on this one lol. Coleman pays $12.99 for 3 hamburgers. How much would Coleman pay for 10 hamburgers? Enter your answe
gayaneshka [121]

Answer:

$43.30

Step-by-step explanation:

What we know,

3 hamburgers = $12.99

Coleman wants to know how much would he have to pay for 10 hamburgers,

In order to find out how much 10 hamburgers cost, you need to find out how much 1 hamburger costs,

To find how much 1 hamburger costs divide 12.99 by 3,

12.99/3

= 4.33

Now we know that 1 hamburger cost $4.33,

Now time to find out how much 10 hamburgers cost,

Multiply 4.33 by 10,

4.33 × 10

= 43.30

Therefore it cost $43.30 for 10 hamburgers.

3 0
2 years ago
Please help me, will give brainliest !!
hoa [83]
4y+25=3y+20
4y+5=3y
***y=5***
so you plug y into both
4(5)+25 and 3(5)+20
45, 100, and 35 are the measurements of the angles.
3 0
3 years ago
NEED HELPS ASAP WILL GIVE BRAINLIST 100 PTS
hodyreva [135]
Hey there!

\ \ \ \ \ \ \ \ \ \Downarrow \ \Downarrow \Downarrow\ \ \ \ \Downarrow \ \ \  \ \Downarrow

\\ \\   \left[\begin{array}{ccc}  \left[\begin{array}{ccc}  \left[\begin{array}{ccc}5(4x + 2)/7y\end{array}\right] \end{array}\right] \end{array}\right]

(5)=(term) \\ \\ (sum)=(+) \\ \\ (product)= (x) \ or \ (*) \\ \\ (4)=(coefficient) \\ \\ (quotient) = (divided \ by \ symbol) \\ \\ (factor)=(2) \\ \\ \boxed{\boxed{Hope \ this \ helps}}

 
4 0
3 years ago
Read 2 more answers
Algebraic models grade 12 math college please write the answers without explaining thank you.
Zinaida [17]

From the problem :

6^4\times6^{-5}

In multiplying expressions with the same bases, the exponent will be added accordingly.

For example :

a^m\times a^n=a^{m+n}

the exponent of a are m and n, and the product will be a raised to the sum of m and n.

Applying this to the problem, we have :

6^4\times6^{-5}=6^{4-5}=6^{-1}

The answer is d. 6^-1

6 0
1 year ago
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