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yawa3891 [41]
3 years ago
5

How much solutions does this equation have?

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

All solutions or infinite.

Step-by-step explanation:

Turn 2y = -4x + 6 into standard form - divide by 2

Y = -2x + 3

2x + y = 3

Substitute the y equation in for y.

2x - 2x + 3 = 3

The x variables cancel out.

3 = 3

Which is true so it means all solutions.

Hope this helps.

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Complete the number pattern 189,209,249,329 809
alexira [117]
189 (+20), 209 (+40), 249 (+80), 329 (+160), 489 (+320), 809 (+640), 1449...

We are adding twice what we added to the previous number. Each time we double what we're adding.
5 0
4 years ago
Soren solves the quadratic equation x^2 + 8x – 9 = 0 using the quadratic formula. In which step did Soren make an error?
pochemuha

Answer:

step 3 didn't divide by 2

Step-by-step explanation:

in step

\frac{ - 8 +  \sqrt{100} }{2}  =  \frac{ - 8 + 10}{2}  = 1 \\  \frac{ - 8 -  \sqrt{100} }{2}  =  \frac{ - 8 - 10}{2}  =  - 9

8 0
4 years ago
Can someone help me please??
denpristay [2]
Yes I can help you what do you need.
3 0
3 years ago
The first four numbers hexa said were 16 32 48 and 64 if she keep counting this way what is the 99th Number hexa will say
rewona [7]

Answer:

316,912,650,057,057,350,374,175,801,344

Step-by-step explanation:

You can see that this sequence is geometric so you need to know the geometric sequence which is x_{n} =r^{n-1} and if you plug in every thing you get x_{n}=2^{99-1} which would be 2^{98} which is 316,912,650,057,057,350,374,175,801,344

6 0
4 years ago
A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If
dsp73

Answer: B) 4.963±0.019.

Step-by-step explanation:

Confidence interval for population mean ( when population standard deviation is not given) is given by :-

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}

, where \overline{x} = Sample  mean

n= Sample size

s= sample standard deviation

t* = critical t-value.

As per given:

n= 50

Degree of freedom = n-1 =49

\overline{x}= 4.963\ lb

s= 0.067 lb

For df = 49 and significance level of 0.05 , the critical two-tailed t-value ( from t-distribution table) is 2.010.

Now , substitute all values in the formula , we get

4.963\pm (2.010)\dfrac{0.067}{\sqrt{50}}\\\\ 4.963\pm (2.010)(0.0094752)\\\\ 4.963\pm0.019045152\approx4.963\pm0.019

Hence,  a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company is 4.963\pm0.019.

Thus , the correct answer is B) 4.963±0.019.

7 0
3 years ago
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