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alukav5142 [94]
3 years ago
12

Solve: 49x2 − 60 = 0

Mathematics
1 answer:
givi [52]3 years ago
3 0

Answer:

C. x = −1.11 and x = 1.11

Step-by-step explanation:

1. I assume you meant 49x^2-60, not 49x2-60.

2. Plug the numbers into the quadratic formula and simplify.

a=49 b=0 c=-60

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48 POINTS ANSWER ASAP​
SVETLANKA909090 [29]

Rewrite 5% as a decimal and then multiply by 60.

X = 0.05 x 60

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What is 54 3/11 as a decimal
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A box of mixed nuts packs weighs 16 1/2 ounces. Each mixed nuts pack weighs 1 3/8 ounces. How many mixed nuts snack packs are in
vfiekz [6]

12 mixed nuts packs

Explanation:

The box weighs 16 1/2 ounces

1 mixed nut = 1 3/8 ounces

let the number of fruit mixed nut in the box = y

1 mixed nut = 11/8 ounces

y = 33/2 ounces

cross multiply:

\begin{gathered} 1(\frac{33}{2})\text{ = y(}\frac{\text{11}}{8}) \\ \frac{33}{2}=\frac{11y}{8} \end{gathered}\begin{gathered} 2(11y)\text{ = 33(8)} \\ 22y\text{ = 33(8)} \\ y\text{ = }\frac{33(8)}{22} \end{gathered}\begin{gathered} y\text{ = 264/22} \\ y\text{ = 12 mixed nuts} \end{gathered}

3 0
1 year ago
G=ta−F3<br> haha its shows GTA
kogti [31]
Hahaha but wait hold up is this an actual question?
3 0
2 years ago
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In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 7.15 km mark at a
ELEN [110]

Answer:

  0.2583 m/s²

Step-by-step explanation:

The relationship between speed, distance, and time is ...

  speed = distance/time

Of course, the relationship between the various units of measure is ...

  • 1 km = 1000 m
  • 1 min = 60 s

The average speed for the first 7.15 km was ...

  (7150 m)/(1500 s) = 4 23/30 m/s

The acceleration is considered to be uniform over the 60-second period. The distance traveled during that time will be the same as the distance traveled at the initial speed for 30 s plus that traveled at the final speed for 30 s. That is, we can compute the final speed as though "Pre" ran 25.5 minutes (1530 s) at his initial speed and the remaining time (133.6 s) at his final speed.

The distance for 25.5 minutes at 4 23/30 m/s is ...

  (1530 s)(143/30 m/s) = 7293 m

The remaining distance is then ...

  10,000 -7,293 = 2,707 . . . meters

And the final speed is ...

  (2707 m)/(133.6 s) ≈ 20.262 m/s

The acceleration is the change in speed divided by the time period:

  (20.262 m/s -4.767 m/s)/(60 s) ≈ 0.2583 m/s²

_____

<em>Comment on these numbers</em>

According to this scenario, Pre's speed for the last 2000 meters, sustained for more than a minute, was more than 20 m/s -- about <em>double the world record</em> speed for a 100 meter sprint. It was a little more than <em>4 times</em> his speed for the first 25 minutes.

5 0
3 years ago
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