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uranmaximum [27]
2 years ago
9

Determine whether the system is consistent, inconsistent, or dependent.

Mathematics
1 answer:
bezimeni [28]2 years ago
5 0
Divide second equation by 2
3x+2y=15
same equation

therefor the system is dependent
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1 4/8 -7/8. Pls help
Travka [436]
12/8-7/8 is 5/8 so the answer is 5/8
4 0
2 years ago
Read 2 more answers
Kylie is 59.54 inches tall. Her friend Adam is 1.37 inch taller than she is. How tall is
SashulF [63]
C; 60.91 inches tall
3 0
1 year ago
Write in slope intercept form<br><br><br> y-(-5)= -4/5(x-5)
vazorg [7]
Well slope Int form is
y= mX +b
So we multiply -4/5 to both "x" and -5

y+5=-4/5x+4
subtract 5 on both sides
y= -4/5x-1
7 0
2 years ago
A patient drinks 5 12ounce cups of coffee daily. The doctor recommends cutting back 60%, how many ounces of coffee can the patie
Dennis_Churaev [7]

The patient can have 204.8 ounce cups of coffee every day.

Step-by-step explanation:

Initial consumption limit of the coffee by the patient- 512-ounce cups

Doctor recommendation-60% cut back in the coffee consumption

Thus according to the doctor's recommendation, if the patient needs to cut back 60% of the coffee consumption, then he is only to consume remaining (100-60)= 40% of his daily consumption.

Total daily consumption he can take= 40% of initial consumption

= (40/100) *512 ounces

= 204.8 ounces of coffee

Hence, the patient needs to take only 204.8 ounces of coffee complying by the Doctor’s recommendation.

5 0
2 years ago
Evaluate each expression for the given value 2/ 5 - 3 1 /2 for k = 15
Ivahew [28]

Answer:

The value of expression for k = 15 is  \frac{13}{5}  

Step-by-step explanation:

Given expression as :

\frac{2}{5} k - 3 \frac{2}{5}  

Or, \frac{2}{5} k - 3  - \frac{2}{5}  

Or, \frac{2}{5} k - \frac{17}{5}  

Now, for k = 15

The expression is

\frac{2}{5} × 15 - \frac{17}{5}  

Or, \frac{30}{5}  - \frac{17}{5}  

Or, 6 - \frac{17}{5}  

Or, \frac{30-17}{5}  

So, \frac{13}{5}  

Hence The value of expression for k = 15 is  \frac{13}{5}   Answer

8 0
2 years ago
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