Determine the zeros of the expression -16t^2-8t+24.

1 answer:

6 0

$-16t^2-8t+24=-8(2t^2+t-3)=-8(2t^2+3t-2t-3)\\\\=-8[t(2t+3)-1(2t+3)]=-8(2t+3)(t-1)\\\\\text{the zeros}\\\\2t+3=0\to t=-1.5\\t-1=0\to t=1\\\\Answer:\ \boxed{-1.5\ and\ 1}$

You might be interested in

Simplify (-2√20k)(5√8k^3) completely

RSB [31]

Answer:

-40 square root of 10k²

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the answer to this

LekaFEV [45]

$4 \dfrac{3}{5} \times \dfrac{5}{12} =$

$= \dfrac{23}{5} \times \dfrac{5}{12}$

You changed the mixed numeral into a fraction correctly.
Now you need to multiply the fractions.
Multiply the numerators together, and multiply the denominators together.

$= \dfrac{23 \times 5}{5 \times 12}$

Now we could multiply out the numerator and denominator, and then we'd need to reduce the fraction, but before we multiply, we see there is a factor of 5 both in the numerator and in the denominator, so we can reduce before we multiply.

$= \dfrac{23 \times 5}{12 \times 5}$

$= \dfrac{23}{12}$

$= 1 \dfrac{11}{12}$

6 0

3 years ago

Simplify to create an equivalent expression. 1+4(6p−9)

myrzilka [38]

Answer: -35+24p

Step by step in picture below

3 0

3 years ago

5 76/10 x 1.75 i need help

alekssr [168]

Answer:

22.05

Step-by-step explanation: