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xenn [34]
2 years ago
13

A rectangle has a length of 6.5m and a width of 7.3m what is the area and the perimeter

Mathematics
2 answers:
Bad White [126]2 years ago
5 0

Step-by-step explanation:

<h3><em><u>Given</u></em><em><u>:</u></em></h3>

Length of the rectangle = 6.5 m

Width of the rectangule = 7.3 m

<h3><em><u>Then</u></em><em><u>:</u></em></h3>

<u>First</u><u> </u><u>case</u><u>,</u>

Area of the rectangle

= length × width

= 6.5 m × 7.3 m

= <em><u>47.45</u></em><em><u> </u></em><em><u>s</u></em><em><u>q</u></em><em><u>.</u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>i</u></em><em><u>)</u></em>

<u>Second</u><u> </u><u>case</u><u>,</u>

Perimeter of the rectangle

= 2(length + width)

= 2(6.5 + 7.3)m

= 2 × 13.8 m

= <em><u>27</u></em><em><u>.</u></em><em><u>6</u></em><em><u> </u></em><em><u>m</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em><em><u>(</u></em><em><u>ii</u></em><em><u>)</u></em>

NeTakaya2 years ago
5 0

Answer:

\pink{\rule{40pt}{555555pt}}

\red,Orange,blue{\rule{40pt}{555555pt}}

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In the diagram, angle 1 = 4x + 30, and angle 3 = 2x +48.
iragen [17]

Answer:

Since we have the information for Angles 1 and 3, and they are vertical, we can set them equal to each other. Once we have done this we can find the measures of them combined, and subtract it from 360 in order to only have the measure of 2 and its vertical angle. Finally, all we need to do now is divide the remaining measure by 2, and this will give us the measure of angle 2.

Angle 1=Angle 3

4x+30=2x+48

2x+30=48

2x=18

x=9

Angle 1=4(9)+30

Angle 1=36+30

Angle 1=66

Angle 1=Angle 3

Angle 1+ Angle 3=132

360-132=228

228/2=114

Angle 2= 114

6 0
2 years ago
Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 feet,
MatroZZZ [7]

The two quadrilaterals are given similar .

In any similar figure sides are in proportion.The second largest side of quadrilateral ABCD is 16 ft .Let the second longest side of quadrilateral EFGH be x ft. These sides will be in proportion to each other .

The second shortest side of quadrilateral EFGH that is GF will be proportional to the third longest side of quadrilateral ABCD that is BC

We can form a proportion with the proportional sides:

\frac{16}{x} =\frac{12}{18}

To solve for x we cross multiply

12x=(16)(18)

12x=288

Dividing both sides by 12 we get

x=24.

The second longest side of quadrilateral EFGH is 24 ft.

6 0
3 years ago
2/5 miles in 1/4 hour what is the speed in miles per hour
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I think its 1/10 miles/hour
8 0
2 years ago
Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

6 0
3 years ago
Read 2 more answers
Please help me with these questions , thank you​
Angelina_Jolie [31]

Answer:

1. 4x^{3} + 26x² + 4x - 48

2. 2x^{3} - 23x² + 60x - 32

3. x^{5} + 7x^{3} - 4x + 2x^{4} + 14x² - 8

4. 2x^{7} + x^{6} - 28x^{5} + 3x + 12

Step-by-step explanation:

To multiply polynomials, simply distribute whatever's outside the largest set of parenthesis, then combine like terms.

1) Distribute the parenthesis (x + 6):

x(4x² + 2x - 8)  +  6(4x² + 2x - 8)

4x^{3} + 2x² -8x  +  6(4x² + 2x -8)

4x^{3} + 2x² -8x + 24x² + 12x - 48

Combine like terms:

4x^{3} + 26x² + 4x - 48

2) Distribute the parenthesis (x - 8):

x(2x² - 7x + 4)  +  (-8)(2x² - 7x + 4)

2x^{3} - 7x² + 4x  +  (-8)(2x² - 7x + 4)

2x^{3} - 7x² + 4x  +  -16x² + 56x - 32

Combine like terms:

2x^{3} - 23x² + 60x - 32

3) Distribute the parenthesis (x + 2):

x(x^{4} + 7x² - 4)  +  2(x^{4} + 7x² - 4)

x^{5} + 7x^{3} - 4x  +  2(x^{4} + 7x² - 4)

x^{5} + 7x^{3} - 4x + 2x^{4} + 14x² - 8

No like terms to combine, so:

x^{5} + 7x^{3} - 4x + 2x^{4} + 14x² - 8

4) Distribute the parenthesis (x + 4):

x(2x^{6} - 7x^{5} + 3)  +  4(2x^{6} - 7x^{5} + 3)

2x^{7} - 7x^{6} + 3x  +  4(2x^{6} - 7x^{5} + 3)

2x^{7} - 7x^{6} + 3x + 8  x^{6} - 28x^{5} + 12

Combine like terms:

2x^{7} + x^{6} - 28x^{5} + 3x + 12

hope this helps!

8 0
2 years ago
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