The pythagorean theroy if a 4=a 8=b what is c

2 answers:

5 0

If a = 4 and b = 8 are the legs and c is a hypotenuse then:

$a^2+b^2=c^2$

Substitute:

$c^2=4^2+8^2\\\\c^2=16+64\\\\c^2=80\to c=\sqrt{80}\\\\c=\sqrt{16\cdot5}\to c=4\sqrt5$

If a = 4 and c are the legs and b = 8 is hypotenuse, then:

 $a^2+c^2=b^2$

Substitute:

 $4^2+c^2=8^2\\\\16+c^2=64\ \ \ |-16\\\\c^2=48\to c=\sqrt{48}\\\\c=\sqrt{16\cdot3}\to c=4\sqrt3$

Minchanka [31]3 years ago

4 0

A^2+b^2=c^2
4^2+8^2=c^2
16+64=c^2
c^2=80
c=the square root of 80 or about 8.9442719......

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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

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So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$