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emmasim [6.3K]
4 years ago
5

Marvin says that all rhombuses are squares. Aretha says that all square are rhombuses. Who is correct? Explain

Mathematics
1 answer:
kati45 [8]4 years ago
6 0
Aretha is correct. Since a rhombus a is a quadrilateral with 4 equal sides in terms of length, (forming 4 right angles in each corner), it can be concluded that all squares are rhombuses. However, only some and not all rhombuses are squares because some rhombuses may have equal lengths but all corners don't form right angles. 

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URGENT!!! ITS DUE IN 5 MINUTES!!!!
gladu [14]

Answer:

d

Step-by-step explanation:

I'll edit my response to tell you why later. it seems like you need this answer now.

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3 years ago
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4x-3y=16 16x+10y=240
Alexeev081 [22]
If you're asking for the solutions to x and y,  x is 10 and y is 8
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What is x if:<br> 9+10=x<br> A. x=109<br> B. x=91<br> C. x=21<br> D. x=19
Vanyuwa [196]
This problem is very simple. All you have to do is add 10 and 9. The correct answer is D. 
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3 years ago
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Let V be the vector space P3[x] of polynomials in x with degree less than 3 and W be the subspace
Pepsi [2]

By definition, the span of two vectors is the set of all possible linear combinations of said vectors. So, any element in W is obtaining by choosing two numbers a,b and building

a(7-8x-8x^2)+b(x^2-(5+6x)) = -8ax^2-8ax+7a+bx^2-6bx-5b

(a)

So, you can show a nonzero polynomial in W by choosing any values of a and b, as long as they're not both zero. To keep it as simple as possible, we can choose for example a=1, b=0 and we obtain one of the base vectors of W:

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Factoring the powers of x, we see that a generic polynomial in W looks like

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So, we must build a polynomial

dx^2+ex+f

such that it is not possible that

\begin{cases}d=b-8a\\e=8a+6b\\f=7a-5b\end{cases}

For example, let's try the polynomial x^2+x+1

We should solve the system

\begin{cases}b-8a=1\\8a+6b=1\\7a-5b=1\end{cases}

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7 0
4 years ago
On 1 May 2012, the cost of 5 grams of gold was 14 000 rupees.
navik [9.2K]

Answer:

51,800 rupees

Step-by-step explanation:

Cost of 5 grams of gold on 1 May 2012 = 14 000 rupees.

The cost of gold decreased by 7.5% from 1 May 2012 to 1 May 2013

Cost of 5 grams of gold on 1 May 2013 = 14,000 - 7.5% of 14,000

= 14,000 - 0.075 × 14,000

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Work out the cost of 20 grams of gold on 1 May 2013

Since, there are four 5 grams of gold in 20 grams of gold

Therefore,

Cost of 20 grams of gold on 1 May 2013 = 12,950 rupees × 4

= 51,800 rupees

4 0
3 years ago
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