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Alekssandra [29.7K]
2 years ago
10

Factorise 4a²_4ab+b²​

Mathematics
1 answer:
Natalija [7]2 years ago
7 0

Answer:

(2a -b)^2

Step-by-step explanation:

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<span>5<span>2–√ i would believe</span></span>
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3 years ago
Order these from least to greatest. ​
soldier1979 [14.2K]

Answer:

order what?

Step-by-step explanation:

4 0
2 years ago
A town's population has been growing linearly. In 2003, the population was 59.000 and the population has been growing by 1,700 p
Rudiy27

Answer:

In 2003, the population was 59000 and the population has been growing by 1,700 people each year.

A.

The equation will be:

59000+1700x = (population 'x' years after 2003)

For x, you plug in the amount of years after 2003.

Like if it is the year 2003, the population is 59000+1700(0)

= 59000

when it is year 2005, the population is 59000+1700(2)

= 62400

B.

The town's population in 2007 will be :

(2007-2003=4)

59000+1700(4)

Population = 65800

C.

59000+1700x=77700

=> 1700x=18700

x = 11

Means 2003+11=2014

Hence, by year 2014 the population will be 77700.

5 0
3 years ago
I NEED HELP!!!!!!!!!!!!!
Igoryamba

Answer:

c+d

Step-by-step explanation:

c as the liters of water in the bucket, and d as the liters of water the jar more.

Together, you have to use addition to answer the question.

Note:

Pls, notify me if my answer is incorrect for the other users that will see this response. Thank you. Have a wonderful rest of your day!

<em>-kiniwih426</em>

4 0
3 years ago
Read 2 more answers
Can someone please explain this to me? Thanks!
Makovka662 [10]

Answer:  Choice D

\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\

==========================================================

Explanation:

Let g(t) be the antiderivative of g'(t) = \sqrt{1+t^3}. We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\

From here, we apply the derivative with respect to x to both sides. Note that the g(\pi) portion is a constant, so g'(\pi) = 0

\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\

\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\

Answer is choice D

5 0
2 years ago
Read 2 more answers
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