Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?
Rewrite the equation −2y=3x+7 in the form
Here the slope of the given line is
If
is the slope of perpendicular line, then

Answer 1: 
Part B. The slope of the line y=−2x+3 is -2. Since
then lines from part A are not parallel to line a.
Since
both lines are not perpendicular to line a.
Answer 2: Neither parallel nor perpendicular to line a
Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then
2·5+5·(-4)=b,
10-20=b,
b=-10.
Answer 3: 2x+5y=-10.
Part D. The slope of the line
is
Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then
7=-4·2+b,
b=7+8,
b=15.
Answer 4: y=-4x+15.
Part E. Consider vectors
These vectors are collinear, then

Answer 5: 
The standard form is:

Degree = 5, leading coefficient=4
The 5th degree polynomial is:
Quintic function
it is a trinomial
<u>What is standard form of a polynomial?</u>
When expressing a polynomial in its standard form, the greatest degree of terms are written first, followed by the next degree, and so on.
So, standard form is:

To find the degree of the polynomial, add up the exponents of each term and select the highest sum ( if there are more than 1 variable in single term) or highest power of variable
Degree = 5
In a polynomial, the leading term is the term with the highest power of x.
So, leading coefficient=4
The 5th degree polynomial is:
Quintic function
It has 3 terms. so, it is a trinomial
To learn more about the standard form of a polynomial from the given link
brainly.com/question/26552651
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Answer:
m=2
Step-by-step explanation:
(0,0) (5,10)
Answer:
it needs to be less dense than water and weight needs to be equal to buoyant force.
Answer: 12.10
Step-by-step explanation:
Given : Mean : 
Standard deviation : 
The formula to calculate the z-score :-

For x= 5 degrees

For x= 10 degrees

The P-value : 

In percent , 
Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%