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2 years ago

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Facts about pasical's education

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Paha777 [63]2 years ago

4 0

Blaise Pascal (June 19, 1623 - August 19, 1662) was a great contributor to math, science, and philosophy, especially Christian philosophy. Interesting Blaise Pascal Facts: Pascal's early education in France was conducted at home by his father due to the prodigious talent and understanding he showed as a child. Did u mean pascal instead of pasical becaus ethere is no such thing. Hope I helped :)

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1/3(22) - 1 -1\2(12)

irinina [24]

The answer is (1/3)...

6 0

3 years ago

Solve this answer<br> 5/3 + 2/5

salantis [7]

5/3+2/5=25/15+6/15=31/15=2 1/15

2 1/15 or 31/15

5 0

3 years ago

Read 2 more answers

The diameter of a circle is 16 kilometers. Find the area of the circle.

Elden [556K]

The diameter of a circle is 16 kilometers. Then the area of the circle is $201.062 \mathrm{km}^{2}$

<u>Solution:</u>

Given that , the diameter of a circle is 16 kilometers

We have to find the area of the circle .

Radius is half of diameter.

$\begin{array}{l}{\text { diameter }=2 \times \text { radius } \rightarrow \text { radius }=\frac{\text {diameter}}{2}} \\\\ {\rightarrow \text { radius }=\frac{16}{2} \rightarrow \text { radius }=8 \text { kilometers. }}\end{array}$

<em><u>The area of circle is given as:</u></em>

$\begin{array}{l}{\text { area of a circle }=\pi \times r^2 \\ {\text { Area of a circle }=\pi \times 8^{2}=\pi \times 64=64 \times 3.14=201.062}\end{array}$

Hence, the area of the circle is 201.062 $km^2$

4 0

3 years ago

Write an equation in point-slope form of the line that passes through (-4,1) and (4,3).

Step2247 [10]

Answer:

An equation in point-slope form of the line that passes through (-4,1) and (4,3) will be:

$y-1=\frac{1}{4}\left(x+4\right)$

Step-by-step explanation:

Given the points

(-4,1)

(4,3)

Finding the slope between the points (-4,1) and (4,3)

$\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(-4,\:1\right),\:\left(x_2,\:y_2\right)=\left(4,\:3\right)$

$m=\frac{3-1}{4-\left(-4\right)}$

Refine

$m=\frac{1}{4}$

Point slope form:

$y-y_1=m\left(x-x_1\right)$

where

m is the slope of the line

(x₁, y₁) is the point

in our case,

m = 1/4

(x₁, y₁) = (-4,1)

substituting the values m = 1/4 and the point (-4,1) in the point slope form of line equation.

$y-y_1=m\left(x-x_1\right)$

$y-1=\frac{1}{4}\left(x-\left(-4\right)\right)$

$y-1=\frac{1}{4}\left(x+4\right)$

Thus, an equation in point-slope form of the line that passes through (-4,1) and (4,3) will be:

$y-1=\frac{1}{4}\left(x+4\right)$

5 0

2 years ago

Divide.<br> a.1 − 2i/2i<br> b. 5 − 2i/5 + 2i<br> c.√3 − 2i/−2 − √3i

Umnica [9.8K]

Answer:

(a) \frac{-1i}{2}-1[/tex]

(b) $\frac{20i+21}{29}$

(c) i

Step-by-step explanation:

We have to perform division

(a) $\frac{1-2i}{2i}$

So after division

$\frac{1-2i}{2i}=\frac{1}{2i}-\frac{2i}{2i}=\frac{-1i}{2}-1$

(b) We have given expression $\frac{5-2i}{5+2i}$

After rationalizing $\frac{5-2i}{5+2i}\times \frac{5-2i}{5-2i}=\frac{(5-2i)^2}{25+4}=\frac{25+4i^2+20i}{29}=\frac{20i+21}{29}$

(c) We have given expression $\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}$

After rationalizing

$\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}\times \frac{-2+\sqrt{3i}}{-2+\sqrt{3i}}=\frac{-2\sqrt{3}+7i+2\sqrt{3}}{7}=\frac{7i}{7}=i$

5 0

3 years ago