The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span> =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
Answer:
A=10506.25 Sq.yards
Step-by-step explanation:
Width of the park=85 yds
Lenght of the park=120 yds
Now,
A=L*W
A=120*85
A=10200 square yards. This is the area of your given rectangle.
Now,
P=2(L+W)
P=2(120+85)
P=2(205)
P=410 this is the perimeter of your given rectangle.
Now,
a=410/4=102.5
Now,calculate the area of newly given distance of each side:
A=a^2
A=102.5^2
A=10506.25 Sq. Yards
This is the new area and is larger than the area of the given rectangle which was 10200 square yards.
Answer:
#9: 1.2
#10: 1.25
Step-by-step explanation:
To find the scale factor of the smaller figure to the larger figure, we're going to be dividing the measurements of corresponding edges.

If you wanted to find the scale factor of the larger figure to the smaller figure, you'd do: 
Question #9:
Left edges:
⇒
= 1.2
Bottom edges:
⇒
= 1.2
<em>(You should get the same number as long as the figures are similar.)</em>
<em />
Question #10:
Bottom edges:
⇒
= 1.25
<em>(There are no corresponding edges with measurements that we can compare.)</em>
<em />
~Hope this helps!~
Answer:
Step-by-step explanation:
Since none of the answer choices match the drawing of the gardener, we assume the question is referring to the drawing of the partner.
The gardener's drawing is 1/4 of actual size. So, in terms of the gardener's drawing, actual size is ...
gardener's drawing = (1/4)actual size
actual size = 4(gardener's drawing)
__
The partner's drawing is 1/20 of actual size, so is ...
partner's drawing = actual size/20 = (4(gardener's drawing))/20
partner's drawing = (4/20)(gardener's drawing)
partner's drawing = (gardener's drawing)/5
__
Then the {length, width} of the partner's drawing are ...
partner's drawing {length, width} = {15 in, 10 in}/5 = {3 in, 2 in}
The partner's drawing has a length of 3 inches and a width of 2 inches.
Following are the solution parts for the given question:
For question A:
In the given question, we calculate
of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

Using the t table we calculate
When
of the confidence interval:
So
confidence interval for the mean weight of shipped homemade candies is between
.
For question B:

Here we need to calculate
confidence interval for the true proportion of all college students who own a car which can be calculated as

Using the Z-table we found
therefore
the confidence interval for the genuine proportion of college students who possess a car is
So
the confidence interval for the genuine proportion of college students who possess a car is between 
For question C:
- In question A, We are
certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams. -
In question B, We are
positive that the true percentage of college students who possess a car is between 0.28 and 0.34.
Learn more about confidence intervals:
brainly.in/question/16329412