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3 years ago

Do u guys have GO MATH 6 gradebooks

Mathematics

1 answer:

swat323 years ago

5 0

Answer:

dude i just told you we arent hackers or whatever. we're students just like you. we find the answers and give them. we dont have books for this stuff

Step-by-step explanation:

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I need help finding this answer because I am pretty sure it’s C but I don’t know

malfutka [58]

Answer: A) -9

Step-by-step explanation:

If x = -2

Then $3^{x}$ = $3^{-2}$

So, it is -9

5 0

2 years ago

Find x and round to the nearest tenth:)

Thepotemich [5.8K]

Answer:

x = 31.8°

Step-by-step explanation:

Applying the Law of Sines, we have:

$\frac{sin(x)}{12} = \frac{sin(75)}{22}$

Multiply both sides by 12

$\frac{sin(x)}{12} \times 12 = \frac{sin(75)}{22} \times 12$

$sin(x) = \frac{sin(75) \times 12}{22}$

$sin(x) = 0.5269$

$x = sin^{-1}(0.5269)$

x = 31.8° (nearest tenth)

8 0

3 years ago

PLZ HELP!

vampirchik [111]

6
9x is rate of function
3x is rate of function
y=mx+b

7 0

3 years ago

April shoots an arrow upward into the air at a speed of 64 feet per second from a platform that is 11 feet high. The height of t

kogti [31]

Answer:

Maximum height of the arrow is 203 feets

Step-by-step explanation:

It is given that,

The height of the arrow as a function of time t is given by :

$h(t)=-16t^2+64t+11$ ..........(1)

t is in seconds

We need to find the maximum height of the arrow. For maximum height differentiating equation (1) wrt t as :

$\dfrac{dh(t)}{dt}=0$

$\dfrac{d(-16t^2+64t+11)}{dt}=0$

$-32t+64=0$

t = 2 seconds

Put the value of t in equation (1) as :

$h(t)=-16(2)^2+64(2)+11$

h(t) = 203 feet

So, the maximum height reached by the arrow is 203 feet. Hence, this is the required solution.

8 0

3 years ago

In mostar, bosnia, the ultimate test of a young man's courage once was to jump off a 400-year-old bridge (now destroyed) into th

Alex787 [66]

To solve this problem, we must assume that the man undergoes constant acceleration as he goes down the river (therefore no other forces must act on him except gravity). Therefore we can use the formula below to calculate for the duration of his fall:

y = y0 + v0 t + 0.5 a t^2

where y is the distance and y0 = 0 since we set the reference point at the bridge, v0 is the initial velocity and is also equal to v0 = 0 since the man started from rest, therefore the equation becomes:

y = 0 + 0 t + 0.5 a t^2

y = 0.5 a t^2

Rewriting in terms of t:

t^2 = 2 y / a

t = sqrt (2y / a)

a is acceleration due to gravity = 9.8 m/s^2

t = sqrt [2 * 23 / 9.8]

t = 2.17 s

Therefore the jump last only about 2.17 seconds.

6 0

3 years ago