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kherson [118]
3 years ago
6

What is the area of this parallelogram? 44 cm² 55 cm² 99 cm² 220 cm² Parallelogram A B C D with side D C parallel to side A B an

d side A D parallel to side B C. Point F is between points D and C and connects to point B with a dotted segment. Point E is between points A and B is connected to point D with a dotted segment. D F B E is a rectangle with all right angles. D F is 4 centimeters. F C is 5 centimeters. E B is 4 centimeters. A E is 5 centimeters. D E is 11 centimeters.

Mathematics
1 answer:
denis23 [38]3 years ago
5 0

Answer:

99 cm²

Step-by-step explanation:

(5 + 4) x 11 = 99

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A tree is 21/2 feet tall.How tall will it be in 3 years if it grows 1/4 foot each year?
r-ruslan [8.4K]

2 1/2 + 3 * (1/4)

2  1/2 + 3/4

2 2/4 + 3/4

 2  5/4

3 1/4 ft

3 0
3 years ago
John has a rectangular picture which is 43 inches long and 27 inches wide. He wants to build a wooden gram for the picture but s
Dvinal [7]

Answer:

The border will be 18.6 inches wide

Step-by-step explanation:

<em>Let the width be w</em>

<em>The length = 43 inches</em>

<em>Total wood = 800 square inches</em>

Formula for area of rectangle = length x width

<em>800 = 43 x w</em>

<em>w = 800/43</em>

w = 18.6 inches

Therefore, the border will be 18.6 inches wide.

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5 0
3 years ago
Read 2 more answers
What is the value of a in the equation 3a + b = 54, when b = 9?
nasty-shy [4]
In order to do this you have to substitute b with 9 and get 3a+9=54, now subtract 9 from both sides to leave a by itself and get 3a=45, now divide 3 in both sides to leave a by itself and get a=15

Hope this helps
4 0
3 years ago
Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
3 years ago
How can I solve for y and what is the answer
Aleks [24]

Answer:

y = 38

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Consider the angles in the large outer triangle, then

x + 46 + 78 = 180

x + 124 = 180 ( subtract 124 from both sides )

x = 56

Consider the 3 angles in the smaller inner triangle, then

y + x + 86 = 180 , that is

y + 56 + 86 = 180

y + 142 = 180 ( subtract 142 from both sides )

y = 38

3 0
3 years ago
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