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tiny-mole [99]
3 years ago
9

Was it evaluated correctly?explain your reasoning.​

Mathematics
1 answer:
NikAS [45]3 years ago
4 0

Answer:

It's not evaluated correctly

Step-by-step explanation:

in the problem 6 × 5 + 30 ÷ 30 we first need to multiply 6 and 5 then divide the 30 by 10 and finally add them up

6×5 = 30

30 ÷ 10 = 3

30 + 3 = 33

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Someone help pls :((((
jasenka [17]

9514 1404 393

Answer:

  y +1 = -3(x -3)

Step-by-step explanation:

The given point is (3, -1), and the given slope is -3. The form suggests you want the point-slope form of the equation for the line:

  y -k = m(x -h) . . . . . . . . line with slope m through point (h, k)

Using the given values, the equation is ...

  y -(-1) = -3(x -3)

  y +1 = -3(x -3)

7 0
3 years ago
Please answer this question !! Thank u tons !! Will give brainliest !!
Anastaziya [24]

Answer: D

Step-by-step explanation:

The key to finding the line perpendicular to the one given is teh slope. The slope is the opposite reciprocal of the original line.

m=3

perpendicular m=-1/3

Now that we know the slope, we can see which of our answer choices have -1/3 as the slope. We can see D is the only option that has -1/3 for slope.

3 0
4 years ago
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Given: Line l is parallel to Line m and Line c is parallel to Line d.
Elanso [62]

Answer:Gn

Step-by-step explanation: bbbbhh

7 0
3 years ago
Liz walks 1/2 mile in 20 minutes on Monday and 3 3/4 in 150 minutes on Sunday is this proportional or non -proportional
ratelena [41]

Answer:

non proportional

Step-by-step explanation:

1/2 doesnt go into 3/4 and 20 doesnt go into 150

4 0
3 years ago
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An algorithm takes 0.5 seconds to run on an input of size 100. How long will it take to run on an input of size 1000 if the algo
dexar [7]

Answer:

linear: 5s

quadratic: 50s

log-linear: 0.75 s

cubic: 500s

Step-by-step explanation:

Let t_1,t_2 be the running time associated with the input of sizes s_1,s_2

If the running time is linear

t_2 = t_1\frac{s_2}{s_1} = 0.5*\frac{1000}{100} = 0.5*10 = 5s

If the running time is quadratic

t_2 = t_1\left(\frac{s_2}{s_1}\right)^2 = 0.5*\left(\frac{1000}{100}\right)^2 = 0.5*10^2 = 50s

If the running time is log-linear

t_2 = t_1\frac{log(s_2)}{log(s_1)} = 0.5*\frac{log(1000)}{log(100)} = 0.5*1.5 = 0.75s

If the running time is cubic:

t_2 = t_1\left(\frac{s_2}{s_1}\right)^3 = 0.5*\left(\frac{1000}{100}\right)^3 = 0.5*10^3 = 500s

6 0
3 years ago
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