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Whitepunk [10]
3 years ago
10

Help me with my math hw Don't do fake answers

Mathematics
1 answer:
anygoal [31]3 years ago
3 0
Its 3 if u multiply
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2x(squared) + 23x-19=0
Helen [10]
Use quadratic formula

for ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}

given
2x^2+23x-19=0
a=2
b=23
c=-19

x=\frac{-23+/- \sqrt{23^2-4(2)(-19)} }{2(2)}
x=\frac{-23+/- \sqrt{529-152} }{4}
x=\frac{-23+/- \sqrt{377} }{4}

x=\frac{-23+ \sqrt{377} }{4} or \frac{-23- \sqrt{377} }{4}
7 0
3 years ago
Number of stickers is less than 1145 but greater than 1100. How many stickers?
dimaraw [331]
45 stickers because 1145 - 1100 = 45.
5 0
3 years ago
This is a timed thing so I need this quick plz
Lady_Fox [76]

The correct answer is a.

6 0
3 years ago
What is the answer to -(5) to the power of two<br>PS Number 6
Damm [24]

Evaluate -(5)^2. Note that Order of Operations rules (PEMDAS) require that we perform exponentiation before multiplication.

Therefore, we evaluate 5^2 first, obtaining 25, and after that multiply this result by -1.


-(5)^2 = -25

6 0
3 years ago
Read 2 more answers
Let n be any natural number greater than 1. Explain why the numbers n! 2, n! 3, n! 4, ..., n! n must all be composite. (This exe
bagirrra123 [75]

Answer:

Because each term of the sequence generates numbers with more than 1 and itself as dividers

Step-by-step explanation:

Just for the sake of correction.

1. Explain\: why\: the\: numbers\: n! +2, n!+ 3, n!+ 4, ..., n! \\n \:must\: all\: be\: composite.

1) Let's consider that

n! =n(n-1)(n-2)(n-3)...

2)And examine some numbers of that sequence above:

n!+2

Every Natural number plugged in n, and added by two will a be an even number not only divisible by two, but in some cases by other numbers for example,n=4, then 4!+2=26 which has four dividers.

3) Similarly, the same happens to

n!+3 and n!+4

Where we can find many dividers.

There's an example of a sequence, let's start with a prime number greater than 1

Let n=11

\left \{ n!+2,n!+3,n!+4,n!+5,n!+6,...n!+n. \right \}\\\left \{ 11!+2,11!+3,11!+4,11!+5,11!+6,11!+7,11!+8,...11!+11 \right \}\\\\

That's a long sequence of consecutive composite numbers, n=11.

\left \{39916802, 39916803,39916804,39916805,39916806,39916807,...,39916811,39916812 \right \}

4 0
3 years ago
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