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Pani-rosa [81]
3 years ago
10

4x=37 a) 3/11 b) 3/28 c) 7/7 d) 12/7

Mathematics
2 answers:
Anettt [7]3 years ago
5 0

4x = 37

divide each side by 4

x = 37/4

4 goes into 37 9 times with 1 left over

x = 9 1/4


Either the problem is incorrect or the answer choices are incorrect

IgorC [24]3 years ago
3 0
It is 37/4, you can divide 4 by 37, so you put it over a fraction, and that is the correct answer!
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A ship tracked for 4 hours heading south and for 5 hours heading west. If the total distance traveled was 191 miles and the ship
Lyrx [107]
Given that the ship was traveling for 4 mph faster heading west than south then let the speed of the ship heading south be x mph, the speed of the ship heading west will be (x+4) mph
the distance is given by:
distance=speed*time
distance heading south was:
x*4=4x miles
distance heading west was:
5(x+4)=5x+20
thus the total distance will be given by:
4x+(5x+20)=191
9x+20=191
9x=191-20
9x=171
x=171/9
x=19 mph
the speed the ship used to travel west was:
x+4
=19+4
=21 mph

7 0
4 years ago
2x - 2y = 22 find the value of y when x equals 2
Vika [28.1K]

Answer:

y=-9

Step-by-step explanation:

If x equals 2, then we can substitute that into the equation, resulting in the equation 2*2-2y=22

We can solve for multiplication first:

4-2y=22

Then we can subtract four on both sides, canceling out the four on the left side:

-2y=18

Now to isolate y, we divide both sides by -2, resulting in the solution:

y=-9

Hope this helps!

6 0
3 years ago
Read 2 more answers
Is a square always sometimes or never a rombus
nataly862011 [7]

Answer:i think it is always a rhombus tbh

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4 0
4 years ago
PLEASE HELP PLEASE HELP PLEASE
Elan Coil [88]

The answer is excepcional i hope it helps

6 0
3 years ago
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An amusement park has 11 roller coasters. In how many ways can you choose 4 of the roller coasters to ride during your visit to
Vadim26 [7]
4!=4*3*2*1=24
11!=11*10*9*8*7*6*5*4*3*2*1=39916800
7!=7*6*5*4*3*2*1=5040
n choose k
how many ways are there to choose k rollercoasters from n choices?

\left(\begin{array}{ccc}n\\k\end{array}\right)=\frac{n!}{k!(n-k)!}

\left(\begin{array}{ccc}11\\4\end{array}\right)=\frac{11!}{4!(11-4)!}=\frac{39916800}{24(7)!}=\frac{39916800}{24(5040)}=\frac{39916800}{120960}=330



330 ways






6 0
4 years ago
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