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Nat2105 [25]
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

Mathematics
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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Find the value of x in the triangle shown below.
Crazy boy [7]

Answer:

i think its the same as one of the angles: 42º

correct me if im wrong

3 0
3 years ago
LXISSUU<br> Simplify<br> -8-1+4.2<br> Enter your answer in the box
jekas [21]

Answer:

-4.8

Step-by-step explanation:

4 0
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
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