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shepuryov [24]
3 years ago
5

What is 2 + 2? Will mark brainliest!!

Mathematics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

It's four.

Step-by-step explanation:

Doesn't require an explanation buuut.... Imagine you have two apples, and a friend gives you two more, you then have four apples.

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Write the equation for this line in slope-intercept form.
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Step-by-step explanation:

The missing value you are looking for is the y-intercept or where the line crosses the y axis.

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Solve the equation.<br> 2.5 = 0 - (-20)<br> 0-17.5<br> 0-22.5<br> O 22.5<br> 0 17.5
Nikolay [14]

Answer:

Step-by-step explanation:Simplifying

0 = 2.5y + -20

Reorder the terms:

0 = -20 + 2.5y

Solving

0 = -20 + 2.5y

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Add '-2.5y' to each side of the equation.

0 + -2.5y = -20 + 2.5y + -2.5y

Remove the zero:

-2.5y = -20 + 2.5y + -2.5y

Combine like terms: 2.5y + -2.5y = 0.0

-2.5y = -20 + 0.0

-2.5y = -20

Divide each side by '-2.5'.

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Simplifying

y = 8

SO the answer is y=8 I think srry if i get it wrong

3 0
3 years ago
The digit 5 appears twice in the number 255,120. How does the total value of the 5 on the right compare to the total value of th
maksim [4K]

Answer:

The value of the first "5" in the number 255,\!120 is ten times that of the second "5\!" in this number.

Step-by-step explanation:

What gives the number "255,\!120" its value? Of course, each of its six digits has contributed. However, their significance are not exactly the same. For example, changing the first \verb!5! to \verb!6! would give 2\mathbf{6}5,\!120 and increase the value of this number by 10,\!000. On the other hand, changing the second \verb!5!\! to \verb!6!\! would give 25\mathbf{6},\!120, which is an increase of only 1,\!000 compared to the original number.

The order of these two digits matter because the number "255,\!120" is written using positional notation. In this notation, the position of each digits gives the digit a unique weight. For example, in 255,\!120\!:

\begin{array}{|r||c|c|c|c|c|c|}\cline{1-7}\verb!Digit!& \verb!2! & \verb!5! & \verb!5! & \verb!1! & \verb!2! & \verb!0!\\\cline{1-7}\textsf{Index} & 5 & 4 & 3 & 2 & 1& 0 \\ \cline{1-7} \textsf{Weight} & 10^{5} & 10^{4} & 10^{3} & 10^{2} & 10^{1} & 10^{0}\\\cline{1-7}\end{array}.

(Note that the index starts at 0 from the right-hand side.)

Using these weights, the value 255,\!120 can be written as the sum:

\begin{aligned}& 255,\!120\\ &= 2 \times 10^{5} + 5 \times 10^{4} + 5 \times 10^{3} + 1 \times 10^{2} + 2 \times 10^{1} + 0 \times 10^{0} \\&=200,\!000 + 50,\!000 + 5,\!000 + 100 + 20 + 0 \end{aligned}.

As seen in this sum, the first "5" contributed 50,\!000 to the total value, while the second "5\!" contributed only 5,\!000.

Hence: The value of the first "5" in the number 255,\!120 is ten times that of the second "5\!" in this number.  

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