HELP PLEASE!

1 answer:

3 0

D.
first, eliminate B and C, because the highest degree in these two equation is 3, a 3-degree equation has 3 roots, not 4.

between A and D, A has the roots 0, 0, 7, and -7, all real roots.

D can be factored into (x²-49)(x²+1)=0
x²=49 or x²=-1
x=7 or -7, x=i or -i two real roots, two unreal roots

for the second equation:
first, find the first term by using the formula:
the nth term=the first term + (n-1)*difference
you are given the 13th term and d=3.7
a13=a1+(13-1)d
1.9=a1+12*3.7
a1=-42.5

the sum of the first 13 terms is S=(a1+a13)*(13/2)=(-42.5+1.9)*6.5=-263.9

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32. Solve the equation below. 12-13 = 17

Helga [31]

12-13
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Therefore the equation is false.

4 0

2 years ago

Read 2 more answers

What sums are equal to 6/12?

nadya68 [22]

Hey there!

"What sums are equal to $\bold{\frac{6}{12}}$ ?"

$\bold{\frac{6}{12}=\frac{1}{2}=0.5 }$

It can't be $\bold{A. \frac{6}{12} + \frac{6}{12}+ \frac{6}{12}+\frac{6}{12}+\frac{6}{12}+\frac{6}{12}}$ because the answer would be $\bold{3}$
It can be $\bold{B.\frac{2}{12} + \frac{1}{12} + \frac{1}{12} +\frac{2}{12}}$
It can also be $\bold{C. \frac{5}{12} + \frac{1}{12}}$
It can also again be $\bold{\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{3}{12}}$
$\boxed{\boxed{\bold{Answers: B,C,D}}}}$