What is the image of (-8, 3) over the x-axis?

Help please! Will give brainly! A rock is dropped from the top floor of a 500-foot tall building. A camera captures the distance

Anna [14]

Answer:

Step-by-step explanation:

1. No it does not as the rock will accelerate downwards (because of gravity) but we can also see that at the first second the rock fell 16 feet compared to the 2 to 3 seconds gap where it fell 80 feet (144-64).

2. All the way down to the ground. Each time gap is getting bigger and bigger (it's accelerating) and the gap between 5 and 6 s should be bigger than 4 to 5 and it is not as the rock cannot fall through the ground but only hit the ground

you may have to change the english a bit and english is not my first language

hope this will help

3 0

3 years ago

What is the slope of the line that passes through the points (-2, 7) and (2,-5)?

Deffense [45]

Answer:

Denote that line: y = ax + b

Line passes (-2, 7) and (2,-5):

=> -2a + b =7

=> 2a + b = -5

Add both side of above equations:

=> 2b = 2

=> b = 1

=> a = (-5 - 1)/2 = -3

=> Slope a = -3

Hope this helps!

:)

4 0

3 years ago

For reviewing purposes

sergeinik [125]

Answer:

(25-16)/9=

=9/9=1 because 5^2 is 25 and 3^2 is 9 (5x5)/(3x3)

5 0

2 years ago

Simplify<br> x2 + 6 − (3x2 − 2x − 5)

Karolina [17]

Answer:

- 2x² + 2x + 11

Step-by-step explanation:

Given

x² + 6 - (3x² - 2x - 5) ← distribute parenthesis by - 1

= x² + 6 - 3x² + 2x + 5 ← collect like terms

= - 2x² + 2x + 11

4 0

3 years ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$