Answer:
0.54 ounces in each tube
Step-by-step explanation:
Add the amount she used so we can add it in. 1.35+0.27= 1.62. Then divide it by the number of tube and since it was a 3 pack you divide it by 3. 1.62/3=0.54 so there is 0.54 ounces in each tube.
We call the ratio between two directly proportional quantities the constant of proportionality. When two quantities are directly proportional, they increase and decrease at the same rate. While these two quantities may increase or decrease, the constant of proportionality always remains the same.
Answer:
undefined
The problem:
Find the slope for the line going through (3,4) and (3,-4).
Step-by-step explanation:
Line up points vertically and subtract.
Then put 2nd difference over 1st.
( 3 , 4)
-(3 , -4)
------------
0 8
So the slope would have been 8/0 but this is undefined.
So the slope is undefined.
Also notice the x's are the same and the y's are difference so this is a vertical line. There is only rise in a vertical line and no run. Recall, slope is rise/run. You cannot divide by 0 so this is why we say the slope is undefined when the x's are always the same no matter the y.
Answer:
Color, favorite pet, genre of music
Step-by-step explanation:
Categorical variables are simply statistical variables which are non - numeric, usually employed in characterization and groupings based on a certain number of fixed attributes, categories. In the options atated above, the categorical variables fall under a heading with a limited and fixed attributes such as color, pet and genre of music. However, options such as number of siblings and profit are purely numeric variables which can take up any numeric digits and allow for direct numeric computation. They are called quantitative variables.
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.
