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2 years ago

Its not a joke please help me

Mathematics

1 answer:

Mariana [72]2 years ago

4 0

Answer:

the same time as a gift card for you to come over and get it done by the time I get home from work and I will be there in a few minutes to get there was a good day for me to be there for you to be there for me to come

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The ratio of the side length of a square to the square's perimeter is always 1 to 4.

makvit [3.9K]

Answer:

Side length of square (inches): 1 ,2, 4, 7

Perimeter of square (inches). : 4, 8, 16, 28

Step-by-step explanation:

Ratio is 1:4 (S:P).

So:

#:8, #=2. This is because 2/8 and 1/4 both = 0.25.

#:16, #=4. 4/16=0.25

7:#, #=28. 4x7=28, 7/28=0.25

Hope that helps

7 0

2 years ago

Need help (midpoints)

Zinaida [17]

Answer:

Step-by-step explanation:

i know it

5 0

3 years ago

Jessica Barth 15 pieces of ribbon each piece of ribbon was 2.25 yards long what is the total amount of ribbon Jessica bu Jessica

Flauer [41]

The answer is 3.375

6 0

3 years ago

ABCD is a trapezium with AB parallel to CD. Find the value of x and the value of y.

Natasha2012 [34]

Answer:

x = 107 , y = 60

Step-by-step explanation:

Each lower base angle is supplementary to the upper base angle on the same side, then

x + 73 = 180 ( subtract 73 from both sides )

x = 107

and

y + 2y = 180

3y = 180 ( divide both sides by 3 )

y = 60

3 0

3 years ago

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

4 0

3 years ago