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1 year ago

(12,-1), (-2,-3)please help

Mathematics

1 answer:

Leviafan [203]1 year ago

4 0

Answer:

13, _1 (12) (-2,3) 1 answer ...by maths 

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Simplify 5т — п — 5п + 2

grigory [225]

Answer:

5t-4n+2

Step-by-step explanation:

n-5n=-4n

5t-4n+2

8 0

3 years ago

Can I have some help with this problem

Rufina [12.5K]

Corresponding measures have the same ratio.
.. (x -1)/4 = (2x +1)/10
.. 5(x -1) = 2(2x +1) . . . . . . multiply by 20
.. 5x -5 = 4x +2 . . . . . . . . eliminate parentheses
.. x = 7 . . . . . . . . . . . . . . . add 5 -4x

x = 7

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3 years ago

Your friend is 1.2 times taller than you. Your friend is 64.5 inches tall. How tall are you?

Andru [333]

You are X, if your friend is 1.2 times taller than you= 1.2X
That's why 64.5=1.2X
X=53.75
But you should write 53.8 inches tall. Because there is one number after point.
As a result, you are 53.8 inches tall

6 0

3 years ago

If ?2000 is placed into a bank account that pays 3% compound interest per year,how much will be in the account after 2 years?

Ipatiy [6.2K]

Answer:

There would be $2,120

Step-by-step explanation:

First figure out what 3% of 2000 is which is 60. Multiply the 60 by 2 to get 120 then add 120 to 2000

3 0

3 years ago

Read 2 more answers

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

(12,-1), (-2,-3)please help​

(12,-1), (-2,-3)please help