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evablogger [386]
3 years ago
12

I have the brainy app on my phone, and I have plus on my phone, but when I try to log in on my computer< it says I still need

to upgrade, how do I use brainy plus on my computer?
Mathematics
2 answers:
IgorC [24]3 years ago
5 0

to use brainy plus on your computer you need to log in with the same account you used on the phone

vladimir2022 [97]3 years ago
3 0

Answer:

try to restart and delete then re download the app:)

Step-by-step explanation:

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GEOMETRY- What is the area of the figure, show your work please!
Kamila [148]
Since this is a 45-45-90 right triangle, the hypotenuse = sqrt2 * leg
24 is the hypotenuse so we need to find the leg.
24 = sqrt2 * leg
Divide both sides by sqrt2
leg = 16.97
A = (1/2)b*h
A = (1/2)*16.97*16.97
A = 144ft^2
8 0
3 years ago
Which line passes through the point (2, -1)
adoni [48]
It’s C because
3*2 - 2* -1 =8
6+2=8
8=8
8 0
2 years ago
Write an expression for the number of small squares in pattern n<br><br> Please properly answer.
antoniya [11.8K]

Answer:

(n + 2)² + 6

Step-by-step explanation:

pattern 1:

Total squares = 15

3² + 6 = 16

Pattern 2:

Total squares = 22

4² + 6

Pattern 3:

Total squares = 31

5² + 6

Looking at the series,

There is an Increment of 1 in squared integer as the pattern increases; there is also a constant integer of 6

Hence, expressing the nth pattern in the form ;

(n + a)² + b

Where n is the nth pattern ; a = (Coefficient of n² in first pattern - 1) = (3 - 1) =2 ;

b = 6

Hence, we have ;

(n + 2)² + 6

3 0
2 years ago
I need help with a word problem .
BigorU [14]
What is the problem ?
4 0
3 years ago
A box contains four 10 nf and eight 100 nf capacitors. (a) what is the probability of obtaining at least two 10 nf capacitors if
grigory [225]

there are 4 of the 10-nf and 8 of the 100-nf capacitors which is a total of 12 items.

the probability of drawing a 10-nf is: 4 out of 12 = \frac{1}{3}

the probability of drawing a 100-nf is: 8 out of 12 =  \frac{2}{3}

(a)  "at least two" means: 2 or 3 or 4

Probability of 2 (10's) and 2 (100's):  \frac{1}{3} x \frac{1}{3} x \frac{2}{3} x \frac{2}{3} = \frac{4}{81}

Probability of 3 (10's) and 1 (100's):  \frac{1}{3} x \frac{1}{3} x \frac{1}{3} x \frac{2}{3} = \frac{2}{81}

Probability of 4 (10's) and 0 (100's):  \frac{1}{3} x \frac{1}{3} x \frac{1}{3} x \frac{1}{3} = \frac{1}{81}

2 or 3 or 4: \frac{4}{81} + \frac{2}{81} + \frac{1}{81} = \frac{7}{81}

(b) without replacement

Probability of 2 (10's) and 2 (100's):  \frac{4}{12} x \frac{3}{11} x \frac{8}{10} x \frac{7}{9} = \frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}

Probability of 3 (10's) and 1 (100's):  \frac{4}{12} x \frac{3}{11} x \frac{2}{10} x \frac{8}{9} = \frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}

Probability of 4 (10's) and 0 (100's):  \frac{4}{12} x \frac{3}{11} x \frac{2}{10} x \frac{1}{9} = \frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}

2 or 3 or 4: \frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9} + \frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9} + \frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9} = \frac{672 + 192 + 24}{12 x 11 x 10 x 9} =  \frac{888}{12 x 11 x 10 x 9}  =  \frac{111}{1485}


8 0
3 years ago
Read 2 more answers
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