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Ket [755]
3 years ago
11

A postcard in the shape of a parallelogram has an area of 12in^2. What are two possible lengths of bases and heights for the pos

tcard
Mathematics
1 answer:
Tpy6a [65]3 years ago
8 0

A postcard is in the shape of a parallelogram. A parallelogram  is a quadrilateral with two pair of parallel sides, opposite sides and opposite angles are equal.

Since, the postcard has an area of 12 square inches.

Since, area of parallelogram = base \times height

As area of parallelogram is 12, it means that the product of base and height is 12 square inches.

So, the possible dimensions of postcard are 3 inches and 4 inches and 2 inches and 6 inches.

So, base = 3 inches , height = 4 inches or base = 4 inches , height = 3 inches.

So, base = 2 inches, height = 6 inches or base = 6 inches , height = 2 inches.

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irina1246 [14]

Answer:

12) A ( 9, 11, 13, 15)

13) A ( o , a)

14) A  (b ∈ ( a, b, c, d, e, f))

15) C ( -364)

16) D (-1°C)

17) B (Negative)

18) C (26 feet)

19) B (7)

Step-by-step explanation:

12) All of these are odd numbers greater than 7

13) These are the only vowels in "MONDAY"

14) Point B cannot belong in a set that already has B

15) -9876 + 9512 = -364

16) 4 - 5 = -1

17) ex: -2 + (-2) = -4

18) -20 + 12 - 18 = -26

19) 3(5) + 4(-2) = 15 - 8 = 7

-Chetan K

8 0
3 years ago
2 points
sattari [20]

Answer:

well the answer is 7 rides

Step-by-step explanation:

25 - 4

=21

21÷3

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(p.s working might not be in the right format but its right)

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ohaa [14]

Answer:0.0057

Step-by-step explanation:

4 0
3 years ago
Is -5/11 equivalent to -5/-11 ? Why or why not?
faltersainse [42]
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3 0
3 years ago
Read 2 more answers
What is the following product? <br>(4x square root 5x^2 + 2^2 square root 6)^2​
tangare [24]

The product is 104 x^{4}+16 \sqrt{30} x^{4}

Explanation:

The given expression is \left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2

Expanding the expression, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)

Now, we shall apply FOIL, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}

Simplifying the terms, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=16 \cdot 5 x^{4}+16 \sqrt{30} x^{4}+4 \cdot 6 x^{4}

Multiplying, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=80 x^{4}+16 \sqrt{30} x^{4}+24 x^{4}

Adding the like terms, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=104 x^{4}+16 \sqrt{30} x^{4}

Thus, the product of the given expression is 104 x^{4}+16 \sqrt{30} x^{4}

7 0
3 years ago
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