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natali 33 [55]
3 years ago
9

Which fraction is closer to 1 than 0

Mathematics
1 answer:
Mrrafil [7]3 years ago
4 0
For some reason, you're not letting us see the choices,
so I can only give you a most general answer:

-- Any positive fraction greater than  1/2  is closer to  1  than it is to  0 .

-- No negative fraction is.
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I’m begging for help!! ASAP
kompoz [17]

This is an awful question.  They mean to say  the number of hours practicing <em>per week</em> (we'll call it h) varies inversely with the <em>time</em> (t) she runs her event.  Unless they mean to imply more practicing makes Tyler slower, in which case she should just get a pizza.

Inverse variation means the product is a constant, we'll call it k.

ht = k

When h = 1 hour t=6 minutes

(1)(6) = k

k = 6  (hour*minutes, but as long as we're consistent we don't need to sweat the unit.)

Decreasing her time by one minute means t = 6 - 1 = 5.  We solve for h.

ht = k

h = k/t = 6/5 = 1.2 hours

Answer: 1.2 hours, second choice

8 0
4 years ago
The figure shown has a parallelogram on top and a rectangle below what is the total area of the figure
stealth61 [152]
Where is the figure ?

4 0
3 years ago
Read 2 more answers
Jon’s test grades are 67,75,73 and 69. What does he need to make on his next test to have a 73 average?
melisa1 [442]

Answer:

he must score 69,75,73and69

Step-by-step explanation:

in order to get average 73

8 0
4 years ago
How do you write this into a number sentence.<br> 32 1/4 divided by 1/8 =258.
Zigmanuir [339]
\frac{32\frac{1}{4}}{\frac{1}{8}}=258
5 0
4 years ago
Use the four-step process to find f'(x) and then find f'(1), f'(2), and f'(3).
Soloha48 [4]

Step 1: evaluate f(x+h) and f(x)

We have

f(x+h) = -(x+h)^2+6(x+h)-5 = -(x^2+2xh+h^2)+6x+6h-5

= -x^2-2xh-h^2+6x+6h-5

And, of course,

f(x)=-x^2+6x-5

Step 2: evaluate f(x+h)-f(x)

f(x+h)-f(x)=-x^2-2xh-h^2+6x+6h-5-(-x^2+6x-5)=-2xh-h^2+6h

Step 3: evaluate (f(x+h)-f(x))/h

\dfrac{f(x+h)-f(x)}{h}=-2x-h+6

Step 4: evaluate the limit of step 3 as h->0

f'(x) = \displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}=-2x+6

So, we have

f'(1) = -2\cdot 1+6 = 4,\quad f'(2) = -2\cdot 2+6 = 2,\quad f'(3) = -2\cdot 3+6 = 0

5 0
3 years ago
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