Answer:
see explanation
Step-by-step explanation:
(a)
Given
2k - 6k² + 4k³ ← factor out 2k from each term
= 2k(1 - 3k + 2k²)
To factor the quadratic
Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)
The factors are - 1 and - 2
Use these factors to split the k- term
1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )
1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term
= (1 - k)(1 - 2k)
1 - 3k + 2k² = (1 - k)(1 - 2k) and
2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)
(b)
Given
2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )
= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term
= (x - 2y)(2a + 3b)
Answer:
h(-9) = -107
Step-by-step explanation:
h(-9) = 4(-9)2 + 3(-9) - 8 = -107
Answer:
<u>y'= 5x^4 + 5^x In(5)</u>
Step-by-step explanation:
<u>Differentiate</u><u> </u><u>with </u><u>Respect</u><u> </u><u>to</u><u> </u><u>x</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>In(</u><u>5</u><u>^</u><u>x</u><u>)</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>x </u><u>In(</u><u>5</u><u>)</u>
<u>with </u><u>respect</u><u> </u><u>to </u><u>x,</u><u> </u><u>we </u><u>have</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>y </u><u>In(</u><u>5</u><u>)</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>5</u><u>^</u><u>x</u><u> </u><u>In(</u><u>5</u><u>)</u>
Hi! The answer is right scalene triangle! Hope this is right! :)
