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3 years ago

Factorise X Squared +13+12

Mathematics

1 answer:

Norma-Jean [14]3 years ago

7 0

Answer:

$\left(x+1\right)\left(x+12\right)$

Step-by-step explanation:

$x^2+13x+12$

$x^2+12x+1x+12$

$x(x+12)+1(x+12)$

$\left(x+1\right)\left(x+12\right)$

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Which is the graph of the linear inequality 2x-3x<12?

lys-0071 [83]

Answer:

Step-by-step explanation:

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3 years ago

Convert 23 m into mm

sergiy2304 [10]

Answer: It is 23,000 Millimeters

Step-by-step explanation:

23 Meters = 23,000 Millimeters

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3 years ago

Read 2 more answers

147 centigrams is equivalent to how many grams

Vladimir79 [104]

147 is equivalent to 1.47 grams

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3 years ago

Need answer asap.thanks

Marrrta [24]

Answer:

$0.46=\frac{23}{50}$

Step-by-step explanation:

To write any decimal as a fraction you divide by 1 and multiply by a number (ranging from 10, 100, 1000 etc.) that will make 0.46 a whole number, this will explain:

Let x = $\frac{0.46}{1}$

10x = $10*\frac{0.46}{1} =\frac{4.6}{10}$

100x = $100*\frac{0.46}{1}=\frac{46}{100}$ this is our perfect fraction, now we simplify later

100x - 10x = $\frac{46}{100} -\frac{4.6}{10}$

90x = $\frac{46}{100} -\frac{4.6}{10} =0$ this is to confirm both fractions are equal

x is the same as $\frac{0.46}{1}$ as $\frac{4.6}{10}$ as $\frac{46}{100}$ but here x = $\frac{46}{100}$ because a fraction has to have no decimals.

So 0.46 is equal any of these values, as a fraction, on the other hand, it's improperly equal to $\frac{46}{100} = \frac{23}{50}$ here I divided by 2 to bring down the proper fraction. (fraction at its simplest form)

6 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago