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trapecia [35]
3 years ago
6

You saved $300 from waiting tables and $100 from babysitting this

Mathematics
1 answer:
Nikolay [14]3 years ago
5 0

86.4

....................

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Use the digits 2, 3, and 5 to create a fraction and a whole number with a product greater than 2
kupik [55]
The fractions which cannbe formed are: 2/3, 3/2, 2/5, 5/2, 3/5 and 5/3.
Now, we find the products with the third number:
2/3*5= 10/3 = 3.33 > 2
3/2*5 = 15 >2= 7.5 > 2
2/5*3= 6/5 (rejected)
5/2*3 = 15/2 > 2
3/5*2= 6/5 (rejected)
5/3*2 = 15/2 = 7.5> 2
5 0
3 years ago
Read 2 more answers
What is the final step for 2y=x-10? Subtract 2 from each side, divide each term by 2, divide each term by -2, or add 2 to each s
Papessa [141]

Answer:

y= x/2 - 5

Step-by-step explanation:

3 0
3 years ago
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Can someone please help me without any links or guessing? I’m putting a lot of points into this one so if you could just help me
riadik2000 [5.3K]

Answer:

C

Step-by-step explanation:

4 0
2 years ago
What is the completely factored form of d4 -8d2 + 16
irinina [24]
According to Vieta's Formulas, if x_1,x_2 are solutions of a given quadratic equation:

ax^2+bx+c=0

Then:

a(x-x_1)(x-x_2) is the completely factored form of ax^2+bx+c.

If choose x=d^2, then:

\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm  \sqrt{64-64} }{2}=4

So, according to Vieta's formula, we can get:

x^2-8x+16=(x-4)(x-4)= (x-4)^2

But x=d^2:

d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2
8 0
3 years ago
Explain WHAT WOULD HAPPEN TO motion IF friction DID NDT EXIST.
MA_775_DIABLO [31]

Answer:

If friction did not exist, it would be hard for us to slow down in running, or putting breaks on a car, and more!

If friction did not exist, your car will slide down the road even if you try your best to put the brakes on a car. Friction is a very important force to slow down things that are moving very fast

Step-by-step explanation:

Hoped this helped.

GeniusUser

5 0
3 years ago
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