PLEASE HELP ILL BRAINLIEST YOU

P is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the

nirvana33 [79]

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: d2=(x−0)^2+(y−0)^2
 =x^2+y^2

To minimize this function d^2 subject to the constraint, 2x+y−10=0
If we substitute, the y-values the distance function can take will be related to the x-values by the line:y=10−2x
You can substitute this in for y in the distance function and take the derivative:
d=sqrt [x2+(10−2x)^2]

d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)
Setting the derivative to zero to find optimal x,
d′=0→10x−40=0→x=4

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).

8 0

3 years ago

A rectangular park is 85 yards wide and 120 yards long.

ki77a [65]

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.

3 0

3 years ago

Find the scale factor of the smaller figure to the larger figure

guapka [62]

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.

$\frac{larger figure}{smaller figure}$

If you wanted to find the scale factor of the larger figure to the smaller figure, you'd do: $\frac{smaller figure}{larger figure}$

Question #9:

Left edges: $\frac{larger figure}{smaller figure}$ ⇒ $\frac{24}{20}$ = 1.2

Bottom edges: $\frac{larger figure}{smaller figure}$ ⇒ $\frac{30}{25}$ = 1.2

(You should get the same number as long as the figures are similar.)

Question #10:

Bottom edges: $\frac{larger figure}{smaller figure}$ ⇒ $\frac{30}{24}$ = 1.25

(There are no corresponding edges with measurements that we can compare.)

~Hope this helps!~

4 0

3 years ago

Could someone help me and explain?

ExtremeBDS [4]

Answer:

length 3 in
width 2 in

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.

6 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ 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

7 0

2 years ago