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3 years ago

(12, 9, 4) What is the area of the parallelogram shown below?

Mathematics

1 answer:

zavuch27 [327]3 years ago

7 0

area of the parallelogram is: b x h

12 x 4 = 48 (d)

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A home is to be built on a 56 foot 9 inch wide lot. The house is 5 feet 5 inches from the side of the lot and is 34 feet 10 inch

daser333 [38]

<span>Total lot - house -other side
=(56 ft 9 in) - (34 ft 10 in) - (5 ft 5 in)
=(55 ft 21 in) - (34 ft 10 in) - (5 ft 5 in)
=16 ft 6 in
Convert to inches: </span>16 ft x 12 in/ft)+ 6 in= Total in inches<span>192 in + 6 in= Total in inches
198 in= Total in inches
Check: One side + house + other side = Lot width
(5 ft 5 in)+(34 ft 10 in)+(16 ft 6 in)=Lot width
56 ft 9 in=Lot width, which equals the given.</span>

8 0

3 years ago

Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?

sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have $V=IR^{2}$ and let $H$ be the subset of $V$ of all points on the line

$4x+3y=12$

We need to find if $H$ is a subspace of the vector space $V$ .

In $IR^{2}$ all the possibilities for own subspace of the vector space $IR^{2}$ are :

$IR^{2}$ itself.

The vector $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$

All lines in $IR^{2}$ that passes through the origin ( $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$ )

We know that $H$ is the subset of $IR^{2}$ of all points on the line $4x+3y=12$

If we look at the equation, the point $\left[\begin{array}{c}0&0\end{array}\right]$ doesn't verify it because :

$4x+3y=12\\4(0)+3(0)=12\\0=12$

Which is an absurd. Therefore, $H$ doesn't contain the origin (and $H$ is a line in $IR^{2}$ ). Finally, it can't be a vector space of $V=IR^{2}$

8 0

3 years ago

Use mental math to find a solution for 2 over eight= 3over x+2

-Dominant- [34]

The Answer: x= −12/7

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME????<br> What is the value of the expression below? A. 8 B. 4 C. 1 D. 2

shutvik [7]

I did this in london! Its B. 4 For londons answers! by mate!

4 0

3 years ago

In the arithmetic sequence {13, 6,-1,-8,...}, what is the common difference?

guapka [62]

<h3>Answer: -7</h3>

Explanation:

Pick any term. Subtract off the previous one to find the common difference.

term2 - term1 = 6-13 = -7
term3 - term2 = -1-6 = -7
term4 - term3 = -8-(-1) = -8+1 = -7

And so on. You only need to pick one of those to show as your steps to your teacher. However, doing all three subtractions is a good way to get practice in seeing how we have an arithmetic sequence. The common difference must be the same each time.

We subtract 7 from each term to get the next term, i.e. we add -7 to each term to get the next one.

6 0

1 year ago