Answer:
A. 4
B. 1
Step-by-step explanation:
The degree of a one-variable polynomial is the largest exponent of the variable.
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<h3>A.</h3>
For f(x) = x^4 -3x^2 +2 and g(x) = 2x^4 -6x^2 +2x -1, the sum f(x) +a·g(x) will be ...
(x^4 -3x^2 +2) +a(2x^4 -6x^2 +2x -1)
= (1 +2a)x^4 +(-3-6a)x^2 +2ax -a
The term with the largest exponent is (1 +2a)x^4, which has degree 4. This term will be non-zero for a ≠ -1/2.
The largest possible degree of f+ag is 4.
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<h3>B.</h3>
The polynomial sum is ...
f+bg = (1 +2b)x^4 +(-3-6b)x^2 +2bx -b
When b = -1/2, the first two terms disappear and the sum becomes ...
f+bg = -x +1/2 . . . . . . a polynomial of degree 1
The smallest possible degree of f+bg is 1.
To arrange the given polynomial in descending order of x, we only look at the exponent of x in each term and then write it in decreasing order of the value. We do as follows:
2xy^4 + 2x^2y − 3y^2 + 10x^3<span>
10x^3 + 2x^2 + 2xy^4 - 3y^2
The degree of the third term would be 5. The degree of a term is the sum of the exponents of the variables involved in the term.
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let's first off take a peek at those values.
let's say the point with those coordinates is point C, so C is 3/10 of the way from A to B.
meaning, we take the segment AB and cut it in 10 equal pieces, AC takes 3 pieces, and CB takes 7 pieces, namely AC and CB are at a 3:7 ratio.
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-4,-8)\qquad B(11,7)\qquad \qquad \stackrel{\textit{ratio from A to B}}{3:7} \\\\\\ \cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em] ~\dotfill\\\\ C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%0A%5C%5C%5C%5C%5C%5C%0AA%28-4%2C-8%29%5Cqquad%20B%2811%2C7%29%5Cqquad%0A%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20A%20to%20B%7D%7D%7B3%3A7%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BA%5Cunderline%7BC%7D%7D%7B%5Cunderline%7BC%7D%20B%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%20%5Ccfrac%7BA%7D%7BB%7D%20%3D%20%5Ccfrac%7B3%7D%7B7%7D%5Cimplies%207A%3D3B%5Cimplies%207%28-4%2C-8%29%3D3%2811%2C7%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0AC%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right) \\\\\\ C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20C%3D%5Cleft%28%5Ccfrac%7B%287%5Ccdot%20-4%29%2B%283%5Ccdot%2011%29%7D%7B3%2B7%7D%5Cquad%20%2C%5Cquad%20%5Ccfrac%7B%287%5Ccdot%20-8%29%2B%283%5Ccdot%207%29%7D%7B3%2B7%7D%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0AC%3D%5Cleft%28%20%5Ccfrac%7B-28%2B33%7D%7B10%7D~~%2C~~%5Ccfrac%7B-56%2B21%7D%7B10%7D%20%5Cright%29%5Cimplies%20C%3D%5Cleft%28%20%5Ccfrac%7B5%7D%7B10%7D~~%2C~~%5Ccfrac%7B-35%7D%7B10%7D%20%5Cright%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20C%3D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D~%2C~-%5Cfrac%7B7%7D%7B2%7D%20%5Cright%29~%5Chfill)
I think ur question is.. If you are making a certain amount u need 2 1/4 flour.. and u need to make half more than that isn't it?
2 1/4= 2.25
Take half of 2. 25= 1.115
Now add 2.25+1.115=3.365
That means you should take 3 cup and a little more than 1/4 cup of flour..
