Expand the following:

Mathematics

2 answers:

liraira [26]2 years ago

6 0

Answer:

8x-16y

Step-by-step explanation:

b.6a/6+18b/6-12c/6

Blababa [14]2 years ago

6 0

Answer:

8x - 16y and a + 3b - 2c

Step-by-step explanation:

(a)

8(x - 2y) ← multiply each term in the parenthesis by 8

= 8x - 16y

(b)

$\frac{6a+18b-12c}{6}$ ← factor out 6 from each term in the numerator

= $\frac{6(a+3b-2c)}{6}$ ← cancel the 6 on numerator/ denominator

= a + 3b - 2c

You might be interested in

Solve for the smallest value of $x$ such that $24x^2 + 17x - 20 = 0.$ Express your answer as a simplified common fraction.

Gala2k [10]

X=5/8 or x=-4/3
here is the work if needed...
24x^2+17x-20=0
Step one: Factor left side of equation.
(8x-5)(3x-4)=0
Step two: Set factor equal to 0.
8x-5=0 or 3x-4=0
x=5/8 or x=-4/3

4 0

2 years ago

in the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect

AlexFokin [52]

Recall the secant-tangent theorem, and you have
EA^2 = EC*CD
12^2 = 8*(x+10)
and now ED = EC+CD = 8+x+10

I suspect a typo somewhere in the murk above

8 0

1 year ago

Write an equation in standard form of the hyperbola described.

marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

$\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}$