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Bess [88]
3 years ago
13

I need help with this is it A. B. C. D.

Mathematics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer: the answer is c hope this helps


Step-by-step explanation:


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How do i find quartiles, 1,2 and 3
Alexus [3.1K]

Lower quartile=25% of total frequency/numbers in the data set

Median=50% of total frequency/numbers in the data set

Upper quartile=75% of total frequency/numbers in the data set

Hope this helps:)

3 0
3 years ago
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What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.
lapo4ka [179]

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1.

Then you can factor this equation:

(t+5)(t+1)=0.

Use the made substitution again:

(x^3+5)(x^3+1)=0.

You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)

and the equation is

(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0.

Here x_1=-\sqrt[3]{5} , x_2=-1 and you can sheck whether quadratic trinomials have real roots:

1. D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}.

2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

6 0
4 years ago
Simplify 24 - 15 - 3+ 2.6.<br> O A. 30<br> O B. 31<br> O C. 126<br> O D. 15
sveticcg [70]

Answer:

D

Step-by-step explanation:

8 0
3 years ago
Identify the type of function represented by f(x)=3(1.5)^x
Murljashka [212]
Here your base (1.5) is greater than zero, so you have exponential growth here.  That " ^ " is the tip-off to exponentiation.

8 0
3 years ago
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I’m not sure with this question help!
labwork [276]

Answer:  C.) 3 to the power of negative two, 1.3 x 10 to the power of negative one, 1/3, and 3 squared.

Step-by-step explanation:

3 to the power of negative 2 = 0.11111111111

3 squared is 1.73205080757

1.3 x 10 to the power of negative 1 = 0.13

1/3 converted to a decimal is 0.333

Now that we know what everything equals we know that anything that doesn't start with 3 squared is right and a little cheat is that only one of them start with 1/3 so we know that the second biggest is 1/3 so it is C.) 3 to the power of negative two, 1.3 x 10 to the power of negative one, 1/3, and 3 squared.

7 0
3 years ago
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