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professor190 [17]
3 years ago
8

george buys 6 apples. Arlene asks if george has the IQ of a fish. He tells her she looks like corned beef hash. Arlene beats Geo

rges fat face in. If Arlene then eats all of Georges apples. How much corned beef hash does Timothy now have?\\  \sqrt{x}  \lim_{n \to \infty} a_n  \sqrt[n]{x}⇔∧⊅║║⊕⊕⊕⊕∵↑↔
Mathematics
1 answer:
user100 [1]3 years ago
5 0
I do not know what you are asking to be solved in this question.
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assume these situations are proportional. write and solve by usinmg a proportion. Everardo paid $1.12 for a dozen eggs at his lo
Bingel [31]
12 eggs for $1.12
3 eggs for $?
As you can see, to get from 12 eggs to 3 eggs, you divide by 4, so do the same with $1.12 to get $0.28
The cost of 3 eggs is $0.28
8 0
3 years ago
If I add 9 to my number,I get six what is my number?
disa [49]
Its tricky naa ans is -3, -3+9=6
7 0
3 years ago
Read 2 more answers
Hulian’s age is three times Thomas’s age. The sum of their age is 36. What is Thomas’s age?
madam [21]
Let t be Thomas, and 3t be Hulian. Then:
t+3t=36
4t=36
t=9
Thomas is 9
☺☺☺☺
4 0
3 years ago
F(x) = 3x* + 4x – 8.<br><br> How many zeros does this have?
Katarina [22]
<h3><u>Question:</u></h3>

f(x) = 3x^2 + 4x – 8.

How many zeros does this have?

<h3><u>Answer:</u></h3>

The given function has two zeros

<h3><u>Solution:</u></h3>

According to the "Fundamental Theorem of Algebra", a polynomial of degree "n"  has  "n"  zeroes

Regardless of odd or even, any polynomial of positive order can have a maximum number of zeros equal to its order. For example, a cubic function can have as many as three zeros, but no more. This is known as the fundamental theorem of algebra.

The degree is the value of the greatest exponent of any term (except the constant ) in the polynomial.

Your function is second degree polynomial, so it has two zeroes.

These can be a mix of rational, irrational, and complex zeroes.

5 0
3 years ago
Discrete Math help please
viva [34]
Set n=1. Then

\displaystyle\sum_{i=1}^1(3i-1)=3(1)-1=2
\dfrac{1(3(1)+1)}2=\dfrac42=2

Both sides match, so the statement holds for this case.

Assume it holds for n=k. Then

\displaystyle\sum_{i=1}^{k+1}(3i-1)=\sum_{i=1}^k(3i-1)+3(k+1)-1
=\dfrac{k(3k+1)}2+3(k+1)-1
=\dfrac{k(3k+1)}2+\dfrac{2(3k+2)}2
=\dfrac{3k^2+7k+4}2
=\dfrac{(k+1)(3k+4)}2
=\dfrac{(k+1)(3(k+1)+1)}2

so that the statement also holds for n=k+1, thus proving the statement by induction.
8 0
3 years ago
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