If I did this right, it should be:
1. 130 degrees
2. 50 degrees
3. 130 degrees
4. 50 degrees
Hope this helped :)
Answer:
Step-by-step explanation:
First, you would have to make it so that T is on one side and everything else is on the other. You would end up with this:
T=1.04+5.60
Solve. You should get to this:
T=6.64
The coefficient of (3y² + 9)5 is <u>15</u>.
A polynomial is of the form a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ₋₁x + aₙ.
Here, x is the variable, aₙ is the constant term, and a₀, a₁, a₂, ..., and aₙ₋₁, are the coefficients.
a₀ is the leading coefficient.
In the question, we are asked to identify the coefficient of (3y² + 9)5.
First, we expand the given expression:
(3y² + 9)5
= 15y² + 45.
Comparing this to the standard form of a polynomial, a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² + ... + aₙ₋₁x + aₙ, we can say that y is the variable, 15 is the coefficient, and 45 is the constant term.
Thus, the coefficient of (3y² + 9)5 is <u>15</u>.
Learn more about the coefficients of a polynomial at
brainly.com/question/9071229
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Answer:
is 0.89
Step-by-step explanation:
in December the average total rainfall in all of the desert together
Answer:
Step-by-step explanation:
So in this example we'll be using the difference of squares which essentially states that: or another way to think of it would be: . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:
Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:
So completely factored form is:
I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: just something that might be useful in some cases.