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zzz [600]
3 years ago
5

Simplify the expression using the order of operations: (21−3)÷32 6th grade math btw

Mathematics
2 answers:
patriot [66]3 years ago
7 0

Answer:

21-3=18

ans /32= 0.5625

amid [387]3 years ago
3 0

Answer:

9/16

Step-by-step explanation:

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Which of the following is equivalent to 2x+3y=6
valentina_108 [34]

Answer:

A. y = -2/3x + 2

Step-by-step explanation:

Your is to isolate the y. The first thing you would do is subtract the 2x,

2x + 3y = 6

-2x        =

3y = -2x +6

then to isolate the ,y, you would next divide everything by 3, so that the ,y, stays by itself.

3y/3 = -2x/3 +6/3

Simplify,

y = -2/3x + 2

5 0
3 years ago
Just need someone to double check if I am correct or not, thanks!
Vsevolod [243]

Answer:

I Think you Got All of them Right but the Last Two

4 0
4 years ago
An open-top box is being designed by cutting a corner piece out of a 16" by 14" piece of metal and folding the sides upwards. Th
Aleksandr-060686 [28]

Answer:

The strip of 16 by 14 inches.

Let x be the corner of the square cut

Then the box would have height as x, length 16-2x and width 14-2x

Hence volume =  

Use derivative to test to find x for maximum volume

V'(x) =  

v"(x) =  

Equate first derivative to 0

Solutions are

x= 2.483 and x = 7.517

Practically cutting more than 7 inches is not possible from 14 inches dimention

Hence 2.483 is the side of square and maximum volume

= 247.508

Step-by-step explanation:

Hope this helps!

7 0
4 years ago
We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0
3 years ago
Algebra 2 help please!
lidiya [134]
V=(6)(x+1)(3x+2) hth
7 0
3 years ago
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