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Pavlova-9 [17]
2 years ago
13

Which is greater: 5 yards or 150 inches?

Mathematics
1 answer:
il63 [147K]2 years ago
4 0

Answer:

5 yards.

Step-by-step explanation:

5 yards = 180 inches

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If I drink decaffeinated coffee, then I do not stay awake.
marishachu [46]

Answer:

the statements are not equivalent

6 0
2 years ago
Ken paid for $12 for two magazines. The cost of each magazine was a multiple of 3. What is Mare the possible prices of the magaz
Artist 52 [7]

Answer:

So the possibilities are

$12 and $ 0    ,    $9 and $ 3      , $6 and $6

Step-by-step explanation:

Given:

Total Magazines = 2

Price = $ 12

To Find:

Possible Prices of magazines = ?

Solution:

As it is given that Prices of magazines is multiples of 3

So

Multiples of 3 up till 12 are  

0, 3 , 6 , 9 , 12

Now

Total Price of magazine is $12

Now The possible prices of the magazines are

First Possibility:

If price of first magazine is $12 then second would be $0 (free)

First possible price $12 and $0

Second Possibility:

If price of first magazine is $9 then second would be $3

Second possible price $3 and $9

Third Possibility:

If price of first magazine is $6 then second would be $6

Third possible price $6 and $6

So the possibilities are

$12 and $ 0    ,    $9 and $ 3      , $6 and $6

8 0
2 years ago
Help me please<br><br><br> A y=9<br><br> B x=-6<br><br> C y= 8x +2
viktelen [127]

Answer:

y = 8x +2 that's the correct answer

7 0
2 years ago
⦁ Find the length of the third angle of a triangle given that the first two angles are 35 and 70. Show your work.
UNO [17]
180 - 70 = 110
110 - 35 = 75
The third angle is 75
3 0
3 years ago
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
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