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2 years ago

I need help on this math question

Mathematics

1 answer:

lina2011 [118]2 years ago

4 0

Answer:

Step-by-step explanation:

7 Sizes * 3 Lengths * 4 Colors = 84

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What is the range of y = sinx in the interval - π ≤ x ≤ 0?

Kaylis [27]

Answer:

$-1 \le y \le 1$

Step-by-step explanation:

Let be $y = \sin x$ , the range of the function corresponds to the set of values of $y$ . The sine function is a bounded function between -1 and 1. Hence, the range of the function is:

$-1 \le y \le 1$

6 0

2 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

2 years ago

ANSWER ASAP<br> THANKS<br> ...............

mylen [45]

Answer:

250

Step-by-step explanation:

5*5=25

25*10=250

+$++$$+$

6 0

3 years ago

Brent has a collection of 84 Bobble Head trophies he needs to box up for the move to his new home. He can fit 7 trophies into on

Alina [70]

$\displaystyle \frac{84\,\text{trophies}}{\left(\frac{7\,\text{trophies}}{\text{box}}\right)}=\frac{84}{7}\,\text{box}\\\\=12\,\text{box}$

Brent will need 12 boxes.

7 0

2 years ago

WILL GIVE BRAINLIEST: What is the local minimum value of the function g(x)=x^4-5x^2+4? (Round answer to the nearest hundredth)

IrinaK [193]

Answer:

Step-by-step explanation:

you can find where the first derivative is 0 to find the critical points

g'(x) = ( x^4 -5x^2 +4)' = 4x³-10x

g'(x) =0, make y =0 to find find where g'(x) is 0

4x³-10x =0 , factor 2x

2x(2x²-5)= 0 , each factor must be 0

2x= 0, so x= 0

2x²-5 =0, so x = ±√5/2

we now have 3 critical points -√5/2, 0, and √5/2

make intervals (-∞, -√5/2), (-√5/2, 0) , (0, √5/2) and (√5/2, +∞)

pick a point to test on each interval: -2, -1, 1 and 2 for example, and

calculate g'(x) = 4x³-10x at those points

for x= -2 we have 4(-2)³-10(-2) = -12 , negative number, decrease

for x= -1 we have 4(-1)³-10(-1) =6, positive number, increase

for x= 1 we have 4(1)³-10(1) = -6, negative number, decrease

for x= 2 we have 4(2)³-10(2) = 12, positive number, increase

we went from a decrease to an increase on intervals (-∞, -√5/2), (-√5/2, 0) so x= - √5/2 ≈ -1.58 is a minimum

we went from a decrease to an increase on intervals (0, √5/2), (√5/2, +∞) so

x= √5/2 ≈ 1.58 is a minimum as well

4 0

2 years ago