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AnnZ [28]
2 years ago
11

How do you factor this problem?

Mathematics
1 answer:
TEA [102]2 years ago
7 0
Factor by grouping.

The answer you're looking for is: 

(3b-7) (b+4)

Hope this helps!
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Probability question. Please answer with work attached.
Helga [31]

Step-by-step Answer:

There is a total of 10 coins, 5 dimes, 2 quarters and three pennies.

By picking a coin, it could be any that shows up out of the 10, so the probability of picking any coin in particular is 1 / 10.

If there are 5 dimes, the probability of picking ANY one particular dime is 1/10, so with 5, the probability of picking ANY of the five dimes is 5/10 = 1/2.

Going along the same line of thought, the probability of picking any of quarters and pennies would be 2/10+3/10 = 5/10 = 1/2 as well.

3 0
3 years ago
What are the real or imaginary solutions of each polynomial equation?
Semmy [17]
1) a
2) c

#1 explanation:

x^4-20x^2=-64
x^4-20x^2+64=0
(find a number that multiplies to the last number aka 64 and adds up to b aka 20)
((x-4)(x-16))^2
x= -2,2,-4,4

same for #2 w different variable and numbers


6 0
3 years ago
Which function is best represented by this graph?
zalisa [80]

Answer:

B- y=\frac{1}{4}x-2

Step-by-step explanation:

To look at the slope you can count over four and then up one and its at the next point.

Also, the place on the y-axis is at -2

So, b makes the most since :)

Hope this helps :) (if it does, brainliest??)

7 0
2 years ago
There's 18 deer and 52 lions if 10 lions eat 17 deer how many deer are left
Mamont248 [21]

Answer:

1 deer

Step-by-step explanation:

18 deer minus 17 eaten leaves 1 deer

7 0
1 year ago
Read 2 more answers
Line E passes through the points (2, 3) and (4, -3). What is the slope of a line perpendicular to line E?
lilavasa [31]
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 2 &,& 3~) 
%  (c,d)
&&(~ 4 &,& -3~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-3}{4-2}\implies \cfrac{-6}{2}\implies \cfrac{-3}{1}

now, a line perpendicular to that one, will have a "negative reciprocal" slope, thus

\bf \textit{perpendicular, negative-reciprocal slope for}\quad \cfrac{-3}{1}\\\\
negative\implies  +\cfrac{3}{ 1}\qquad reciprocal\implies +\cfrac{ 1}{3}\implies \cfrac{1}{3}
7 0
2 years ago
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