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Virty [35]
3 years ago
15

help me with measurement please! if a square can become a rectangle can a rectangle become a square?

Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0
<span>Actually, every square is a rectangle, since the angles in a square are always right angles. That's more than saying that a square can be a rectangle; it is one.
</span><span>
And since squares are rectangles, you know that some rectangles are squares--namely, the squares are!
</span>
<span>We teach children "this is a rectangle, that is a square. The rectangle's sides are different lengths". But when you grow up, it's important to think of the square as a SPECIAL rectangle, because it is all that a rectangle is, and more.
</span>
A square is still a rectangle, but it's not just a rectangle, it's a (pedigreed) square.
In summary, yes a rectangle can be a square.

Hope this helped :)
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Need answer ASAP will give brainliest check out the picture
valentinak56 [21]

Answer:

i think... x=6

Step-by-step explanation:

3x+9=x+21

3x−x=12

2x=12

x=6

8 0
2 years ago
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Verdich [7]

Answer:

x=-1,\:x=-7,\:x=i,\:x=-i

Step-by-step explanation:

Considering the equation

x^4+8x^3+8x^2+8x+7=0

Solving

x^4+8x^3+8x^2+8x+7

\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)

As

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=7,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

Solving

\frac{x^4+8x^3+8x^2+8x+7}{x+1}

=x^3+7x^2+x+7

Putting \frac{x^4+8x^3+8x^2+8x+7}{x+1} =  x^3+7x^2+x+7 in equation [A]

So,

\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

=\left(x+1\right)x^3+7x^2+x+7

As

x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)

So,

Equation [A] becomes

=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)

So,  the polynomial equation becomes

\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\mathrm{Solve\:}\:x+1=0:\quad x=-1

\mathrm{Solve\:}\:x+7=0:\quad x=-7

\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i

\mathrm{The\:solutions\:are}

x=-1,\:x=-7,\:x=i,\:x=-i

Keywords: polynomial equation

Learn polynomial equation from brainly.com/question/12240569

#learnwithBrainly

5 0
3 years ago
Read 2 more answers
Factor completely, 9x4 -225y8
Furkat [3]

Step-by-step explanation:

9x4 -225y8 \\  = 3 \times 3 \times x \times 2 \times 2 -3 \times 5 \times 3 \times 5 \times y \times 2 \times 2 \times 2

4 0
3 years ago
We select a random sample of 25 observations from a normally distributed population with an unknown population variance. the com
MrMuchimi
Sorry, I do not know the answer to your question
4 0
3 years ago
I need help!! #28 #29<br> Please!!!!!
Marta_Voda [28]

Answer:

Step-by-step explanation:

We would apply the formula for binomial distribution. It is expressed as

P(x = r) = nCr × q^(n - r) × p^r

Where

n = number of samples

p = probability of success.

q = probability of failure

From the information given,

n = 12

p = 90% = 90/100 = 0.9

q = 1 - p = 1 - 0.9 = 0.1

28) Probability that at least 10 are ripe within 4 days is expressed as

P(x ≥ 10) = P(x = 10) + P(x = 11) + P(x = 12)

P(x = 10) = 12C10 × 0.1^(12 - 10) × 0.9^10 = 0.23

P(x = 11) = 12C11 × 0.1^(12 - 11) × 0.9^11 = 0.38

P(x = 12) = 12C12 × 0.1^(12 - 12) × 0.9^12 = 0.28

P(x ≥ 10) = 0.23 + 0.38 + 0.28 = 0.89

29) Probability that no more than 9 are ripe within 4 days is expressed as

P(x ≤ 9) = 1 - P(x ≥ 10)

P(x ≤ 9) = 1 - 0.89 = 0.11

4 0
3 years ago
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