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3 years ago

Use the quadratic formula to solve x^2-2x+2=0

1 answer:

3 0

The quadratic formula is...
$\frac{-b (+/-)\sqrt{b^2-4ac}}{2a}$
your a will be in front of your x^2, b in front of your x, and c will be the number without an x. Plug those into the equation and you'll get two answers. That will be it

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How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits a

kiruha [24]

Answer: There are 1400 different combinations.

Step-by-step explanation:

The conditions are:

we have 4-digits: abcd.

all the digits are different.

a is an odd number, and b is an even number.

Then, for a, we have the options 1, 3, 5, 7 and 9 (so we have 5 options).

for b, we have the options 0, 2, 4, 6 and 8 (so we have 5 options).

for c, we can have odd or even numbers, so we have 8 options ( remember that there where 2 numbers already taken away, this is why we have only 8 options).

for d we have 7 options (because 3 numbers are already taken).

Then the number of combinations is equal to the product of the number of options for each selection:

C = 5*5*8*7 = 1400

3 0

3 years ago

<br />How do I solve 2x+y=10

ololo11 [35]

X is 3 y is 4 hope this helps :)

7 0

3 years ago

Read 2 more answers

Please help im desperate

gogolik [260]

Answer:

∠

∘

and

∠

113

∘

Explanation:

By what is given,

∠

∘

...

(

)

, and,

∠

180

∘

...

(

)

Since,

∠

∘

, by (1), we get,

∠

∘

−

∘

Using this in

(

)

, we get,

∠

180

∘

−

∘

113

∘

3 0

2 years ago

The fifth through the eighth terms of a geometric sequence are given by a5 = 80, a6 = 160, a7 = 320, a8 = 640. Which formula can

Ganezh [65]

hlo I am ram and my frn is shyam how are you I am also fine here

3 0

2 years ago

The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?

olga nikolaevna [1]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\$ $\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
% (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2$

3 0

3 years ago