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lana66690 [7]
3 years ago
10

What is 6P4? 15 24 360 720

Mathematics
2 answers:
Lunna [17]3 years ago
8 0
It is 360 I’m 99.99% sure
ASHA 777 [7]3 years ago
4 0

Answer:

360

Step-by-step explanation:

I just done the quiz and got it right :) Choice C! Have a great day! Sub to my YT Channel plz

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Factor the polynomial
Art [367]

Find the GCF (Greatest Common Factor)

GCF = 2x

Factor out the GCF (Write the GCF first and then, in parentheses, divide each term by the GCF.)

2x (2x^3/2x + 4x^2/2x + 6x/2x)

Simplify each term in parentheses

-2x(x^2 + 2x + 3)

<u>Answer D. -2(x^2 + 2x + 3)</u>

7 0
3 years ago
What is 625 x 625 Then what is the atomic weight of hydrogen.
Illusion [34]

Answer:

la respuesta es:

Step-by-step explanatio:

625 x 625

390,625

6 0
3 years ago
July average temperature: 78°F,January average temperature: 52°F What is the percentage difference?
ycow [4]
I’m guessing you subtract 78-52 then divide by original which was 52.. so July is 50% warmer.
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3 years ago
find the interest earned on an account with $450 principal, and interest rate of 6% over a period of 8 years
puteri [66]
 
                 8
450(1+.06)  = 717.23
4 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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