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Slav-nsk [51]
3 years ago
8

PLEASE HELP ILL BRAINLIEST YOU

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
7 0

Answer:

R = (-2,0)

E = (2,2)

F = (4,-4)

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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: <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).
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

<em>(You should get the same number as long as the figures are similar.)</em>

<em />

Question #10:

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

<em>(There are no corresponding edges with measurements that we can compare.)</em>

<em />

~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
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