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Darina [25.2K]
3 years ago
14

Find g(x), where g(x) is the translation 4 units down of f(x)=x2. Write your answer in the form a(x–h)2+k, where a, h, and k are

integers. g(x)=
Mathematics
1 answer:
kirill [66]3 years ago
7 0

Answer: g(x) = (x+3)^2

The answer should be in the form of a(x-h)2 + k.

Since we are not translating up or down, the k value is 0. We will not be adding any values to the outputs of the original equation here. a will equal 1 since no stretches are occurring. So we are left with the h value.

f(x) = x2 is given.

g(x) = a horizontal translation of 3 units left. This means that the h value is -3, because in order to move left, you need to take 3 away from all x values (inputs). So the point (2,4) on f(x) would become (-1,4) on g(x).

How do we write this?

g(x) = f(x-h) where h = -3.

So this is g(x) = f(x - (-3)) = f(x+3)

g(x) = (x+3)^2

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32.8

Step-by-step explanation:

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What is the nth term?
kumpel [21]

Let a_k denote the <em>k</em>th term of the sequence. Then

a_k=a_1+d(k-1)

where <em>d</em> is the common difference between consecutive terms in the sequence and <em>a</em>₁ is the first term.

The sum of the first <em>n</em> terms is

S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n

From the formula for a_k, we get

S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)

S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k

S_n=na_1+\dfrac{d(n-1)n}2

S_n=\dfrac n2(2a_1-d+dn)

So we have d=-5, and 2a_1-d=16 so that a_1=\frac{11}2.

Then the <em>n</em>th term in the sequence is

a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}

7 0
4 years ago
PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS
Reil [10]

Given

x+1 = \sqrt{7x+15}

We have to set the restraint

x+1\geq 0 \iff x \geq -1

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}

So, we have to impose

x-7\geq 0 \iff x \geq 7

Squaring both sides, we have

(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0
3 years ago
Get the general term for the sequence being your t3 = 11 and the t20 = 244.2
marusya05 [52]

Answer:

nth term = t_{n} = 7.639(1.2)^{n - 1}

Step-by-step explanation:

Let us assume that the given sequence is a G.P.

Now, if the first term of the G.P. is a and the common ratio is r, then

Third term = t_{3} = ar^{2} = 11 .......... (1) and  

20th term = t_{20} = ar^{19} = 244.2 ........... (2)

Now, dividing equation (2) with equation (1) we get

\frac{ar^{19} }{ar^{2} } = \frac{244.2}{11} = 22.2

⇒ r^{17} = 22.2

⇒ r = 1.2.

Hence, from equation (1) we get

a(1.2)² = 11

⇒ a = 7.639 (Approx.)

Therefore, the general term of the sequence i.e. nth term = t_{n} = 7.639(1.2)^{n - 1} (Answer)

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