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3 years ago

[2x+5]=3x+4 what is the solution

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

4 0

Answer:

x=-1

Step-by-step explanation:

2x+5=3x+4

-4 -4

2x+1=3x

-2x -2x

1=-x

1/-1 =-x/-1

-1=x

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The area of a square is 16 cm2. The square is the base of a cube.

choli [55]

the answer is 96 cm ^ 2

16 × 6 =96

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3 years ago

Hi please answer and you dont NEED to show work but if u do i will give brainliest

Advocard [28]

Answer:

$y=-\frac{2}{5} -1$

Step-by-step explanation:

So first thing is that slope intercept form is setup like this

y=mx+b

m=slope

b=y intercept

the slope is change in y over change in x

as y goes down 2 x goes up 5

so the slope is -2/5

the y intercept is at (0,-1) so the y intercept is -1

therefore the answer is

$y=-\frac{2}{5} -1$

have a great day :)

8 0

3 years ago

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It took Mary 40 minutes to run 8 laps around the playground, if she spent the same time running each lap, how long would it take

Sergeeva-Olga [200]

Answer:

it should take her 45 minutes to run 9 laps

Step-by-step explanation:

40/8 = 5

9*5 = 45

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3 years ago

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Solve for x check your solution

abruzzese [7]

Area for a triangle is (1/2) bh or bh/2
So using this info, b = 1.5 m and h = x and A = 1.5 m^2

A = bh/2
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8 0

3 years ago

The deck of a bridge is suspended 275 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of

Ivenika [448]

Answer:

Our equation for the height is:

y(t) = 275 - 16*t^2.

a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:

$AV = \frac{y(t2) - y(t1)}{t2 - t1}$

Then:

i) t1 = 4s, t2 = 4s + 0.1s = 4.1s

The average velocity is:

$AV = \frac{(275 - 16*4.1^2) - (275 - 16*4^2)}{4.1 - 4} = \frac{16(4^2 - 4.1^2)}{0.1} = -129.6$

And the units will be ft/s, so the average speed is:

-129.6 ft/s

The minus sign is because te pebble is falling down.

ii) t1 = 4s, t2 = 4s + 0.05s = 4.05s

The average velocity is:

$AV = \frac{(275 - 16*4.05^2) - (275 - 16*4^2)}{4.05 - 4} = \frac{16(4^2 - 4.05^2)}{0.05} = -128.8$

So the average speed is -128.9 ft/s

iii) t1 = 4s, t2 = 4s + 0.01s = 4.01s

The average speed is:

$AV = \frac{(275 - 16*4.01^2) - (275 - 16*4^2)}{4.01 - 4} = \frac{16(4^2 - 4.01^2)}{0.01} = -128.16$

The average speed is -128.16 ft/s.

b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.

v(t) = dy(t)/dt.

v(t) = -2*16*t + 0

Then the velocity at t = 4s is:

v(4s) = -32*4 = -128

The instantaneous velocity at t = 4s is -128 ft/s.

3 0

3 years ago