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Aleksandr-060686 [28]
4 years ago
10

Need help 1-4 I'll appreciate it

Mathematics
1 answer:
uranmaximum [27]4 years ago
6 0
For Q1 add 135 and 45 together. If it comes out to be 180 which it does the angles are supplementary. In Q2 add 23 and 67 together and if it comes out to 90 it is complementary, but if it doesn't it is neither. In Q3 take 3x+45=180 subtract 45 on both sides so you get 3x=135 then divide by 3 to get X. Your answer should be 45. For Q4 look that exact question up and find a good answer for it. Hope this could help
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Step-by-step explanation:

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Read 2 more answers
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
Help please ?<br> its for a quiz and i do not really understand the problem.
Margarita [4]

Answer: I'm not entirely sure but i think its the last one! How this helps!

Step-by-step explanation:

I followed the information on the graph and the one with a 7.2 circumference had a 22.66 cm diameter!

4 0
3 years ago
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