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2 years ago

A rectangular cracker has a length of 5 centimeters and an area f 20 square centimeters. Find its perimeter.

Mathematics

1 answer:

iren [92.7K]2 years ago

4 0

Answer:

18cm

Step-by-step explanation:

We have the length of the rectangle but we need to find the width which we'll call w. Area is a, l is length.

a = l * w

20 = 5 * w

divide both sides by 5 to cancel the multiplication by 5 on the right side

4 = w

The rectangle is going to have two sides that are the length of l, and two sides that are the the length of w. We're looking for the perimiter p.

p = 2l + 2w

p = (2 * 5) + (2 * 4)

p = 18

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Point M is located in the third quadrant of the coordinate plane, as shown.

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Answer:

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Step-by-step explanation:

The points of M means there x and y will both be negatives.

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3 years ago

Find the midpoint of the line segment defined by the points: (5, 4) and (−2, 1) (2.5, 1.5) (3.5, 2.5) (1.5, 2.5) (3.5, 1.5)

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Answer:

$\boxed {\boxed {\sf (1.5 , 2.5)}}$

Step-by-step explanation:

The midpoint is the point that bisects a line segment or divides it into 2 equal halves. The formula is essentially finding the average of the 2 points.

$(\frac {x_1+x_2}{2}, \frac {y_1+ y_2}{2})$

In this formula, (x₁, y₁) and (x₂, y₂) are the 2 endpoints of the line segment. For this problem, these are (5,4 ) and (-2, 1).

x₁= 5
y₁= 4
x₂= -2
y₂= 1

Substitute these values into the formula.

$( \frac {5+ -2}{2}, \frac {4+1}{2})$

Solve the numerators.

5+ -2 = 5-2 = 3
4+1 = 5

$( \frac {3}{2}, \frac{5}{2})$

Convert the fractions to decimals.

$(1.5, 2.5)$

The midpoint of the line segment is (1.5 , 2.5)

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3 years ago

The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is .

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Answer:

(-21,-19)

$\sqrt{849}$

Standard form

Step-by-step explanation:

We are given the equation of circle

$x^2+y^2+42x+38y-47=0$

General equation of circle:

$x^2+y^2+2gx+2fy+c=0$

Centre: (-g,-f)

Radius: $\sqrt{g^2+f^2-c}$

Compare the equation to find f, g and c from the equation

$g\rightarrow 21$

$f\rightarrow 19$

$c\rightarrow -47$

Centre: (-21,-19)

Radius (r) $=\sqrt{21^2+19^2+47}=\sqrt{849}$

Standard form of circle:

$(x+21)^2+(y+19)^2=849$

The centre of circle at the point (-21,-19) and its radius is $\sqrt{849}$ .

The general form of the equation of a circle that has the same radius as the above circle is standard form.

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