Factor completely 3ab − 5ax + 9b − 15x.

Mathematics

2 answers:

lora16 [44]3 years ago

7 0

Answer:

$(a+3)(3b-5x)$

Step-by-step explanation:

We are given that an expression

$3ab-5ax+9b-15x$

We have to find the factors of given expression

Rearrange the terms of given expression then we get

$3ab+9b-5ax-15x$

Now, we taking common 3b from first two terms and -5 x from last two terms

$3b(a+3)-5x(a+3)$

Now. taking common a+3 then we get

$(a+3)(3b-5x)$

victus00 [196]3 years ago

3 0

<span>(3b - 5x)(a +3) is the answer

</span>

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A golf ball is hit from the ground with an initial velocity of 208 ft./s. assume the starting height of the ball is 0 feet. how

V125BC [204]

Check the picture below, it hits the ground at y = 0

thus

$\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}\\
----------\\
v_o=208\\
h_o=0
\end{cases}$

$\bf h(t)=-16t^2+208t+0\implies h(t)=-16t^2+208t\impliedby 
\begin{array}{llll}
\textit{now let's}\\
make\\ h(t)=0
\end{array}
\\\\\\
0=-16t^2+208t\implies 0=t(208-16t)
\\\\\\
\begin{cases}
0=t\\
----------\\
0=208-16t\\
\qquad 16t=208\\
\qquad t=\frac{208}{16}
\end{cases}$

and I'm sure you know how much that is

it hits it at two points, as you notice in the picture, at 0 seconds, when it's first hit and goes up, and when it falls down.