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3 years ago

11

HELP WILLING TO CASHAPP if answered quick and correctly + 100 points The points plotted below on the graph of a polynomial which

of the following X values is the best approximation of a root of the polynomial? Check all that apply
X=-0.21
X=-4.6
X=-2.3
X=3
X=1
X=-3.5

2 answers:

Ghella [55]3 years ago

6 0

Answer:

Step-by-step explanation:

Did you ever find out the answer????

Mashutka [201]3 years ago

3 0

first 1

4th one

i think sorry if its wrong

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LXISSUU<br> Simplify<br> -8-1+4.2<br> Enter your answer in the box

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-4.8

Step-by-step explanation:

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3 years ago

Write the terms a 1a1, a 2a2, a 3a3, and a 4a4 of the following sequence. If the sequence appears to converge, make a conjec

Arturiano [62]

Answer:

Step-by-step explanation:

WE are given that $a_n = \frac{(-1)^{n+1}}{5n-4}$ . Then, to now the first for terms, we must replace n by 1,2,3,4 respectively. Then

$a_1 = \frac{(-1)^2}{5(1)-4} = \frac{1}{1}= 1$

$a_2 = \frac{(-1)^3}{5(2)-4} = \frac{-1}{6}$

$a_3 = \frac{(-1)^4}{5(3)-4} = \frac{1}{11}= 1$

$a_4 = \frac{(-1)^5}{5(4)-4} = \frac{-1}{16}= 1$

Note that as n increase, $a_n$ gets closer to 0. So, the limit of this sequence is 0.

6 0

3 years ago

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

3 years ago