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3 years ago

9

Which picture illustrates the prophase stage of mitosis?

2 answers:

posledela3 years ago

7 0

The answer is B. where nuclear membranes form around each set of chromosomes resulting in two nuclei.

netineya [11]3 years ago

3 0

Prophase is when the nuclear envelope is starting to break down and the chromatin is condensing. The picture that illustrates this is D.

For reference...
Picture A - Anaphase
Picture B - Telophase
Picture C - Metaphase
Picture D - Prophase

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Name the algebraic property demonstrated in the example below.

lilavasa [31]

The algebraic property demonstrated in the example below is Transitive Property of Equality. There we can see how the first thing is equal to the second one and notice that the first one is equal to the third one too. This is a Transitive Property of Equality in a nutshel.

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3 years ago

Help ASAP being timed please.

Yanka [14]

Answer:

Its the third one

Step-by-step explanation:

Plot all of the points onto the graph, and draw a line through all of the points. It most closely resembles the third one

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3 years ago

What is the answer for 5-2+7x6-2?

sergiy2304 [10]

Answer: 43

Step-by-step explanation:

5-2+7x6-2

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3 years ago

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aivan3 [116]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Simplify the expression state any excluded values<br> 2a^2-4a+2<br> ---------------<br> 3a^2-3

chubhunter [2.5K]

Answer:

The simplified form is $\dfrac{2(x-1)}{3(x+1)}$ .

$x =1$ is the excluded value for the given expression.

Step-by-step explanation:

Given:

The expression given is:

$\dfrac{2a^2-4a+2}{3a^2-3}$

Let us simplify the numerator and denominator separately.

The numerator is given as $2a^2-4a+2$

2 is a common factor in all the three terms. So, we factor it out. This gives,

$=2(a^2-2a+1)$

Now, $a^2-2a+1=(a-1)(a-1)$

Therefore, the numerator becomes $2(a-1)(a-1)$

The denominator is given as: $3a^2-3$

Factoring out 3, we get

$3(a^2-1)$

Now, $a^2-1$ is of the form $a^2-b^2=(a-b)(a+b)$

So, $a^2-1=(a-1)(a+1)$

Therefore, the denominator becomes $3(a-1)(a+1)$

Now, the given expression is simplified to:

$\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}$

There is $(x-1)$ in the numerator and denominator. We can cancel them only if $x\ne1$ as for $x=1$ , the given expression is undefined.

Now, cancelling the like terms considering $x\ne1$ , we get:

$\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}$

Therefore, the simplified form is $\dfrac{2(x-1)}{3(x+1)}$

The simplification is true only if $x\ne1$ . So, $x =1$ is the excluded value for the given expression.

8 0

3 years ago