What is the answer for a+b-c?

I WILL MAKE YOU BRAINLEST

Strike441 [17]

It is the vendor that sells 1.5 L for $2.70

8 0

2 years ago

, list the first five terms of each sequence. f(n + 1) = −2f(n) + 8 and f(1) = 1 for n ≥ 1

svet-max [94.6K]

Answer: 1, 7, 1, 7, 1.

Step-by-step explanation:

The given function : $f(n+1)=-2f(n)+8$ and f(1) = 1 for n ≥ 1

We put value of n=1,2,3,4 , we get

For n=1 , $f(2)=f(1+1)=-2f(1)+8=-1(1)+8=-1+8=7$

For n=2 , $f(3)=f(2+1)=-2f(2)+8=-1(7)+8=1$

For n=3 , $f(4)=f(3+1)=-2f(3)+8=-1(1)+8=7$

For n=4 , $f(5)=f(4+1)=-2f(4)+8=-1(7)+8=1$

Hence, the first five terms of the given sequence are 1, 7, 1, 7, 1.

7 0

3 years ago

Please help! Could you answer all!?

Brums [2.3K]

Answer:

1.) 48

2.) 65

3.) 36

Step-by-step explanation:

1.) If the equation is 6(x-4) and x = 12, then all we have to do is plug in the value of x. When we plug in, all we do is substitute 12 for x because they mentioned in the question that x = 12. So, we end up getting 6(12 - 4). After solving this, we get 48.

2.) This problem is a lot like the last problem. All we need to do is substitute /plug in the values of x and y into the equation, to get 4(4^2) - 35/7 - (8 + 14). After solving, we get 65.

3.) . This problem, once again, is also a lot like the last problems. We need to substitute the value of x into the equation 8x+12. Since we know from the problem that x is 3, all we have to do is 8 * 3 + 12.

3 0

2 years ago

Read 2 more answers

Solve each equation (quadratic pattern)

dusya [7]

Answer: x = 2

Step-by-step explanation:

$2^{2x}-2^x-12=0\\\\\text{Let u = }2^x\\\\u^2-u-12=0\\(u-4)(u+3)=0\\\\u-4=0\quad and\quad u+3=0\\u=4\qquad and\quad u=-3\\\\\text{Substitute u with }2^x\\2^x=4\qquad and \quad 2^x=-3\\2^x=2^2\quad and\quad \text{not possible}\\\boxed{x=2}$

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Answer: x = 0

Step-by-step explanation:

$3^{2x}+3^{x+1}-4=0\\\\3^{2x}+3^x\cdot3^1-4=0\\\\\text{Let u = }3^x\\u^2+3u-4=0\\\\(u+4)(u-1)=0\\\\u+4=0\quad and\quad u-1=0\\u=-4\qquad and\quad u=1\\\\\text{Substitute u with }3^x\\3^x=-4\qquad and\quad 3^x=1\\\text{not possible}\ and\quad 3^x=3^0\\.\qquad \qquad \qquad \qquad \boxed{x=0}$

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Answer: No Solution

Step-by-step explanation:

$4^x+6\cdot 2^x+8=0\\\\2\cdot 2^x+6\cdot 2^x+8=0\\\\\text{Let u = }2^x\\2u+6u+8=0\\8u+8=0\\8u=-8\\u=-1\\\\\text{Substitute u with }2^x\\2^x=-1\\\text{not possible}$