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LUCKY_DIMON [66]
3 years ago
12

Help pls I need it someone

Mathematics
1 answer:
Anika [276]3 years ago
7 0

Answer:

search up demos graphing calculator and plug in the numbers there.

Step-by-step explanation:

its super easy to use try it :)

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Hanna bought 5 lottery cards with $25. How much is each card worth?​
stepladder [879]
Each card is worth $5
8 0
3 years ago
Read 2 more answers
Solve for x<br> 4(x + 1) = –28
lora16 [44]

Answer:

X=-8

Step-by-step explanation:

4x+4=-32

Add -4 to both sides

4x=-32

decide both sides by 4

x=-8

4 0
2 years ago
Out of 600 seniors at a local high school, 60% went on the senior trip. At the hotel, one room was reserved for every 4 students
butalik [34]

Given:

Total number of senior students = 600

60% went on the senior trip.

One room was reserved for every 4 students.

To find:

The total number of reserved rooms.

Solution:

60% went on the senior trip from total 600 students. So, number of students who went on tripe is

600\times \dfrac{60}{100}=360

Now, one room was reserved for every 4 students. So,

\text{Required number of rooms}=\dfrac{\text{Number of students who went on tripe}}{\text{Number of students in room}}

\text{Required number of rooms}=\dfrac{360}{4}

\text{Required number of rooms}=90

Therefore, the required number of reserved rooms were 90.

4 0
3 years ago
Monique bought 5 1/2pounds of chicken for $18.51. What was the cost per pound?
UkoKoshka [18]
About 3.40 rounded to the tenths place
5 0
2 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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