Note that h(x) = 4x^2 - 8x - 60 factors to 4(x^2 - 2x - 15), and that the roots (zeros) of (x^2 - 2x - 15) are -5 and +3. Thus, the smallest zero is -5.

3 0

3 years ago

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Oliver read for 450450450 minutes this month. His goal is to read for 10\, percent more minutes next month. If Oliver meets his

dybincka [34]

Step-by-step explanation:

450450450=100%

? =110%

110×450450450/100

495495495 mins=2nd month

add

495495495+450450450=

945945945 minutes

7 0

2 years ago

Simplify the expression csc(-x)/1+tan^2x)

Charra [1.4K]

Assuming ya meant $\frac{csc(-x)}{1+tan^2(x)}$

to slimplify, we use a variation of the pythagorean identity and a decomposition into the sin and cos

for the pythaogreaon identity
$cos^2(x)+sin^2(x)=1$
divide both sides by $cos^2(x)$
$1+tan^2(x)=sec^2(x)$ since $\frac{sin(x)}{cos(x)}=tan(x)$
subsitute
$\frac{csc(x)}{sec^2(x)}$

recall that $csc(x)=\frac{1}{cos(x)}$
also that cos(x) is an even function and thus cos(-x)=cos(x)
therfore $csc(-x)=\frac{1}{cos(-x)}=\frac{1}{cos(x)}=csc(x)$
so we get

$\frac{csc(x)}{sec^2(x)}$
decompose them into $\frac{1}{cos(x)}$ and $\frac{1}{sin^2(x)}$ to get $\frac{\frac{1}{cos(x)}}{\frac{1}{sin^2(x)}}$
multiply by $\frac{sin^2(x)}{sin^2(x)}$ to get
$\frac{sin^2(x)}{cos(x)}$
we can furthur simlify to get
$(\frac{sin(x)}{cos(x)})(sin(x))=tan(x)sin(x)$
the expression simplifies to tan(x)sin(x)

5 0

3 years ago

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