Find the value of the variable. If the answer is not a whole number, round to the nearest tenth.

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

Two secants drawn to a circle from a common external point, then

The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is

5(5 + x) = 4(4 + 10) = 4 × 14 = 56

25 + 5x = 56 ( subtract 25 from both sides )

5x = 31 ( divide both sides by 5 )

x = 6.2 → C

You might be interested in

What is the quotient: (6x4 + 15x3 + 10x2 + 10x + 4) ÷ (3x2 + 2)?

Kryger [21]

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

$\frac{\left(6x^4+15x^3+10x^2+10x+4\right)}{\left(3x^2+2\right)}$

Now, we will find the quotient by factoring the numerator:

$\mathrm{Use\:the\:rational\:root\:theorem}\\a_0=4,\:\quad a_n=6\\\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \\\mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2,\:3,\:6\\\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1,\:2,\:3,\:6}\\\\-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2\\\\=\left(x+2\right)\frac{6x^4+15x^3+10x^2+10x+4}{x+2}\\\\=\frac{6x^4+15x^3+10x^2+10x+4}{x+2}=6x^3+3x^2+4x+2\\\\$

Now, we will factor it again:

$=\left(6x^3+3x^2\right)+\left(4x+2\right)\\\\=2\left(2x+1\right)+3x^2\left(2x+1\right)\\\\=\left(2x+1\right)\left(3x^2+2\right)$

At last we get our factorised form :

$=\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)\\\\=\frac{\left(x+2\right)\left(2x+1\right)\left(3x^2+2\right)}{3x^2+2}\\\\=\left(x+2\right)\left(2x+1\right)\\\\=2x^2+5x+2$