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svetlana [45]
2 years ago
15

Hey please help where it has the < thing it says “ones” “tens” “nines”

Mathematics
1 answer:
Keith_Richards [23]2 years ago
5 0

Answer:

9 \times 9 = 10nines - 1nine \\  9 \times 9 = 90 - 9 \\ 9 \times 9 = 81 \\ thank \: you

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0.3 km = ? mi. rounded to the nearest hundred​
Sergio [31]

Answer:

0.19

Step-by-step explanation:

There are approximately .62 miles in a kilometer

6 0
3 years ago
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Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0
2 years ago
.............helppppppppppppppppppp
raketka [301]

Answer:

what????????????????

8 0
3 years ago
Read 2 more answers
Find the 5th term of the sequence in which T1=8 and Tn=3t n-1
yan [13]
T_n = 3 * T_(n-1)

Long way (always works!)
T_5 = 3*T_4,
T_4 = 3*T_3
T_3 = 3*T_2
T_2 = 3*T_1

T_5 = 3*3*3*3*T_1 = 81*T_1 = 81*8 = 648!

Short way (sometimes it works!)

T_n = 3^(n-1) * T_1 (this case is a geometric series of ratio-=3)
T_5 = 3^4*8 = 648
6 0
2 years ago
What is the ratio of circles to total shapes
Soloha48 [4]
Hm i think it might be 2:8?
6 0
2 years ago
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