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2 years ago

Find all the zeros of the equation. Need help finding the zeros.

1 answer:

6 0

Given the expression
-3x^4+27x^2+1200=0
let x^2=a
we can re-write our expression as:
-3a^2+27a+1200=0
-3(a^2-9a-400)=0
a^2-9a-400=0
factorizing the above we have:
a^2+16a-25a-400=0
a(a+16)-25(a+16)=0
(a+16)(a-25)
thus replacing back x^2 we have:
(x^2+16)(x^2-25)
=(x^2+16)(x-5)(x+5)
factorizing (x^2+16) we get
x^2=+/-√-16
x=+/-4i
thus the zeros of the expression are:
x=-5, x=5 , x=-4i, x=4i

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3/10=15/50=18/? Para ya pLis

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The answer is 18/10 .

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3 years ago

Need help on this if anyone can

Lunna [17]

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2 years ago

36 jelly donuts is 200 percent of how many jelly donuts?

Art [367]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

Please explain each step 5x-7=7x-17

EleoNora [17]

Let's get the same terms on the same side.
I am going to remove 5x from both sides of the equation by subtracting 5x

5x - 5x -7 = 7x -5x - 17
-7 = 2x - 17 now lets remove the 17 from both sides
-7 + 17 = 2x (why do we add 17? Think of 2x -17 as 2x + -17. Tor remove a negative number we add it.
Now we finally have 10=2x and if we divide both sides by 2

10 = 2x
2 2 we get that x = 5
Let's check by substituting 5 in the original equation:
5(5) - 7 = 7(5) - 17
25 - 7 = 35 - 17
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6 0

3 years ago

In arithmetic sequence tn find: S10, if t2=5 and t8=15

makkiz [27]

Given:

$\bold{t_2=5}\\\\\bold{t_8=15}\\\\$

To find:

$\bold{T_n=?}\\\\\bold{S_{10}=?}$

Solution:

$\to \bold{t_2=5}\\\\ \bold{a+d=5} \\\\ \bold{a=5-d} ................(i)\\\\ \to \bold{t_8=15}\\\\ \bold{a+ 7d=15}.................(ii)\\\\$

Putting the equation (i) value into equation (ii):

$\to \bold{5-d+7d=15}\\\\\to \bold{5+6d=15}\\\\\to \bold{6d=15-5}\\\\\to \bold{6d=10}\\\\\to \bold{d=\frac{10}{6}}\\\\\to \bold{d=\frac{5}{3}}\\\\$

Putting the value of (d) into equation (i):

$\to \bold{a=5-\frac{5}{3}}\\\\\to \bold{a=\frac{15-5}{3}}\\\\\to \bold{a=\frac{10}{3}}\\\\$

Calculating the $\bold{t_n :}$

$\to \bold{t_n=a+(n-1)d}\\\\\to \bold{t_n=\frac{10}{3}+(n-1)\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}+\frac{5}{3}n-\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}-\frac{5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{10-5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{5}{3} +\frac{5}{3}n}\\\\ \to \bold{t_n=\frac{5}{3}(1+n)}\\\\$

Formula:

$\bold{S_n = \frac{n}{2}[2a + (n - 1)d ] }\\$

Calculating the $\bold{S_{10}} :$

$\to \bold{S_{10} = \frac{10}{2}[2\frac{10}{3} + (10- 1)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + (9)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 9 \times \frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 3\times 5 ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 15 ] }\\\\\to \bold{S_{10} = 5[\frac{20+ 45}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{65}{3} ] }\\\\\to \bold{S_{10} = 5 \times \frac{65}{3} }\\\\\to \bold{S_{10} = \frac{325}{3} }\\\\\to \bold{S_{10} =108.33 }$

Therefore, the final answer is " $\bold{\frac{5}{3}(1+n)\ and\ 108.33}$ "

Learn more:

brainly.com/question/11853909

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3 years ago