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kotykmax [81]
3 years ago
10

Given: If a parallelogram has 4 congruent sides, then it is a rhombus. In Parallelogram ABCD, AB = BC = CD = DA What can you con

clude?
Mathematics
2 answers:
kakasveta [241]3 years ago
7 0
It is identical in form ... 4 equal sides therefore 4 congruent sides therefore a rhombus?
irina [24]3 years ago
7 0
The answer is B)ABCD is a rhombus

You can conclude that ABCD is a rhombus<span>. Since the lengths of the sides are equal, we know that they are also congruent. You can use the given information to conclude that </span>Parallelogram<span> ABCD is a </span>rhombus<span>.</span>
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If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
2 years ago
What is the value of z in the equation 2z + 6 = –4? (5 points) <br> 5 <br> 1 <br> –1 <br> –5
PtichkaEL [24]
The answer is -5. Hope this helps.
6 0
3 years ago
Read 2 more answers
Solve 4x + 1 = 3 – 2x
romanna [79]

Answer:

x = 1/3

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

4x + 1 = 3 - 2x

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Add 2x on both sides:                    6x + 1 = 3
  2. Subtract 1 on both sides:                6x = 2
  3. Divide 6 on both sides:                   x = 1/3

<u>Step 3: Check</u>

<em>Plug in x to verify it's a solution.</em>

  1. Substitute:                    4(1/3) + 1 = 3 - 2(1/3)
  2. Multiply:                        4/3 + 1 = 3 - 2/3
  3. Add/Subtract:               7/3 = 7/3

Here we see that 7/3 does indeed equal 7/3.

∴ x = 1/3 is a solution of the equation.

7 0
3 years ago
Read 2 more answers
Please help me I don’t understand
ioda

Answer:

-8 I believe

Step-by-step explanation:

8 0
2 years ago
Question number 8. Prove that 3 + √5 is an irrational number.​
Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

\\  \sf \: 3 +  \sqrt{5}  =  \frac{a}{b}   \\  \\   \qquad \: \tiny \sf{(where \:  \: a \:  \: and \:  \: b \:  \: are \:  \: integers \:  \: and \:  \: b \:  \neq \: 0)} \\

\\  \sf \:  \sqrt{5}  =  \frac{a}{b}  - 3 =  \frac{a - 3b}{b}  \\

Since, a, b and 3 are integers. So,

\\ \sf \:  \frac{p - 3b}{b}  \\  \\  \qquad \tiny \sf{ \: (is \:  \: a \:  \: rational \:  \: number \:) } \\

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

\\

Hence, 3 + √5 is an irrational number.

7 0
3 years ago
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