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4 years ago

How many Solutions are there to the question below

Mathematics

2 answers:

Natalija [7]4 years ago

6 0

I think is B because any value of X makes the equation true

Nadusha1986 [10]4 years ago

3 0

Answer:

Step-by-step explanation:

By expanding we can get

5x+7x+48=12x+48

Simplifying gets us

12x+48=48+12x

12x=12x

This is true for any number so therefore there are infinity solutions.

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A cardboard box 6 units high, 4 units wide, and 3 units long can hold 80 pounds of brown rice. A second cardboard box has the sa

iren [92.7K]

I hope this helps but use lingith × with × higth so you would × 6 ×4 ×3

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3 years ago

I'll give the usual stuff of brainliest and thanks to the first person, plus some extra points.

tino4ka555 [31]

Answer:

I really don't get your question

6 0

3 years ago

Dana has 4 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the ra

Sedaia [141]

Answer:

2hrs 40 minutes

Step-by-step explanation:

The formula for distance is expressed as:

$d=st, s=speed, t=time$

#The time taken to run:

$d=st\\\\t=d/s\\\\t_r=d/8$

#The time taken to walk:

$d=st\\\\t=d/s\\\\t_w=d/2$

#Since the distances are the same:

$d_r=8t_r,d_w=2t_w\\\\8t_r=2t_w\\\\t_r+t_w=4->t_w=4-t_r\\\\\therefore8t_r=2(4-t_r)\\\\8t_r=8-2t_r\\\\t_r=1\frac{1}{3} \hrs\\\\t_w=4-1\frac{1}{3}=2\frac{2}{3}\ hrs$

Hence, Dana spends 2hrs 40 minutes walking back

8 0

3 years ago

Which of the following equations is the equation 2 - 3y + 6x = 0 written in slope-intercept form?

IgorLugansk [536]

The answer us c
..........

7 0

4 years ago

Read 2 more answers

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

4 years ago