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3 years ago

Solve for x please need help need answer ASAP

Mathematics

1 answer:

Solnce55 [7]3 years ago

3 0

From the given diagram , in smaller triangle BCD , Using Pythagoras theorem we can write

$x^2=p^2+4^2\\\\x^2=p^2+16\\\\$

In Triangle ABD , we can write

$q^2=21^2+p^2\\\\$

In Triangle ABC

$(21+4)^2=q^2+x^2\\\\q^2+x^2=625\\$

Now using the above three equations we can write

$21^2+p^2+x^2=625\\\\21^2+x^2-16+x^2=625, x^2=p^2+16\\\\2x^2+425=625\\\\2x^2=625-425\\\\2x^2=200\\\\x^2=100\\\\x=10\\$

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What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

So,

$b=0,b-4=0$

$b=0,b=4$

Therefore, the solutions are:

$b=0,\:b=4$

4 0

3 years ago

Read 2 more answers

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Find out more on equation at: brainly.com/question/2972832

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2 years ago

The lines y1 = 2x - 6, y2 = -3x + 4, y3 = -1/2x + 4 are the sides of a right triangle. Find the perimeter and area of the triang

sergiy2304 [10]

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