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Dmitriy789 [7]
3 years ago
13

17.The speed limit of a highway is 55 miles per hour. A car is traveling at least 65 miles per hour. How many miles per hour m o

ver the speed limit is the car traveling?
55 – m ≥ 65; m ≥ 10 m + 65 ≤ 55; m ≤–10 65 – m ≤ 55; m ≤ 10 55 + m ≥ 65; m ≥ 10**** 18. Chris earns $7.00 per hour working on the weekends. He needs at least $210.00 for a new cell phone. How many hours, h, does Chris need to work on the weekends to buy a new phone?
h/7.00 > 210.00; h> 30; 30 hours h/7.00 < 210.00; h < 30; 30 hours 7.00h≥ 210.00; h≥ 30; 30 hours** 7.00h≤ 210.00; h ≤ 30; 30 hours
Mathematics
1 answer:
liraira [26]3 years ago
3 0
17. 55 + m ≥ 55

18. 210 / 7 = 30 h 

h ≥ 30 hours



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Karo-lina-s [1.5K]

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5 0
3 years ago
16. 75/100 whats inside the big 1
slava [35]

answer: 15

there is 15 left.

6 0
3 years ago
Find the limit (enter 'DNE' if the limit does not exist)
vaieri [72.5K]

Answer:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)

first we'll rationalize our expression:

\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}

\sqrt{-2x^2-6y^2+1}+1

this is our simplified expression, now we can apply our limits:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)

\sqrt{-2(0)^2-6(0)^2+1}+1

1+1

2

the limit does exists and it is 2.

5 0
3 years ago
What is the 10th term of the geometric sequence 400, 200, 100...?
Vsevolod [243]

ANSWER

a_ {10} = \frac{25}{32}

EXPLANATION

The given geometric sequence is

400, 200, 100...

The first term is

a_1=400

The common ratio is

r =  \frac{200}{400}  =  \frac{1}{2}

The nth term is

a_n=a_1( {r}^{n - 1} )

We substitute the known values to get;

a_n=400(  \frac{1}{2} )^{n - 1}

a_ {10} =400(  \frac{1}{2} )^{10 - 1}

a_ {10} =400(  \frac{1}{2} )^{9}

a_ {10} = \frac{25}{32}

4 0
3 years ago
Hello there looking for help here thanks. giving brainliest if able to
Zolol [24]

Answer:

angles 2 and 10 are corresponding Angles while angles 6 and 11 are alternate Angles

5 0
3 years ago
Read 2 more answers
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