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wel
2 years ago
15

B%207%29" id="TexFormula1" title=" \tt(6x4−2x3 +7x2 −x) + ( 7x4+ 2x3−5x + 7)" alt=" \tt(6x4−2x3 +7x2 −x) + ( 7x4+ 2x3−5x + 7)" align="absmiddle" class="latex-formula">
\\

With Solution!!​​
Mathematics
2 answers:
Arturiano [62]2 years ago
5 0

Answer:

13(x)⁴ + 7(x)²-6(x) + 7

Explanation:

\hookrightarrow \sf (6x^4 -2x^3+7x^2-x)+(7x^4+2x^3-5x+7)

<u>remove the parenthesis</u>

\sf \hookrightarrow 6x^4-2x^3+7x^2-x+7x^4+2x^3-5x+7

<u>group the variables</u>

\hookrightarrow \sf 6x^4+7x^4-2x^3+2x^3+7x^2-x-5x+7

<u>add similar terms</u>

<u />\sf  \hookrightarrow  13x^4+7x^2-x-5x+7

<u>simplify</u>

\hookrightarrow \sf 13x^4+7x^2-6x+7

Vadim26 [7]2 years ago
3 0

\qquad\qquad\huge\underline{{\sf Answer}}

Let's solve ~

\qquad \tt \dashrightarrow \:\tt(6x {}^{4} −2x {}^{3}  +7x {}^{2} −x) + ( 7x {}^{4} + 2x {}^{3} −5x + 7)

\qquad \tt \dashrightarrow \:\tt6x {}^{4} −2x {}^{3}  +7x {}^{2} −x +  7x {}^{4} + 2x {}^{3} −5x + 7

Now, move all like terms aside :

\qquad \tt \dashrightarrow \:6 {x}^{4}  + 7 {}x^{4}  - 2 {x}^{3}  + 2 {x}^{3}  + 7 {x}^{2}  - x - 5x + 7

Add the like terms ~

\qquad \tt \dashrightarrow \:13 {}x^{4}   + 7 {x}^{2}   + 6x + 7

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navik [9.2K]

Answer:

(4, -6)

Step-by-step explanation:

A reflection across the y-axis means the point is equal but opposite distance from the y-axis. It begins at 4 units to the LEFT (negative) of the y-axis, so after the reflection, it should be 4 units the the RIGHT (positive) of the y-axis. This has no change on the y-value of the point, because no matter the y-value, the point will still be the same distance from the y-axis. Long story short, if you're reflecting across the y-axis, change the sign of the x-coordinate. If you're reflecting across the x- axis, change the sign of the y-coordinate. That's why the answer is (4, -6).

6 0
3 years ago
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One whole jar= 3/3 or 1

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convert to improper fractions

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ANSWER: 3 3/4 ounces can make a full jar.

Hope this helps! :)
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Which shows the equation of the line using function notation?
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In a straight horizontal line the slope is always 0. This cancels out the 'x' factor so you get f(x)=b.

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