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3 years ago

What is a the area of parallelogram RSTU? Square units

Mathematics

1 answer:

gtnhenbr [62]3 years ago

4 0

Answer: 32 units²

Step-by-step explanation:

1. Circumscribe a rectangle on the parallelogram.

2. You can see that four triangles are formed. So, to calculate the area of the parallelogram you must subtract the area of four triangles from the area of the rectangle.

3. The triangles SVR and UYT are equal, therefore their areas are equal. The triangles UXR and SWT are equal, therefore their areas are equal.

4. So, the area of the parallelogram is:

$A_p=(10unit)(6units)-2(\frac{6*2}{2})-2(\frac{4*4}{2})=32units^{2}$

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The graph of F(x), shown below, resembles the graph of G(x) x2, but it has

Bingel [31]

Correct answer is: F(x)=-x^2-3

Step-by-step explanation:

We are given a function:

The graph of is also shown in the given question figure.

It is a parabola with vertex at (0,0).

Sign of is positive, that is why the parabola opens up.

General equation of parabola is given as:

Here, in G(x), a = 1

Vertex (h,k) is (0,0).

As seen from the question figure,

The graph of F(x) opens down that is why it will have:

Sign of as negative. i.e.

And vertex is at (0,-3)

Putting the values of a and vertex coordinates,

Hence, the equation of parabola will become:

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2 years ago

Factorise y^2 - 16<br> (y squared)

cricket20 [7]

$y^2-16=y^2-4^2=(y-4)(y+4)\\\\\text{used}\ a^2-b^2=(a-b)(a+b)$

8 0

2 years ago

Read 2 more answers

Given the functions k(x) = 5x − 8 and p(x) = x − 4, solve k[p(x)] and select the correct answer below.

Mazyrski [523]

P(x) = x-4

k(x-4) = 5(x-4) - 8 = 5x - 28

So the answer is 2

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3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

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3 years ago

What is the slope of a line that passes through the points (-3,-6) and (8, 16) ?

denis-greek [22]

Answer:

slope = 2

Step-by-step explanation:

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2 years ago