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3 years ago

10

Determine whether the systems have one solution, no solution or infinitely solutions

1 answer:

larisa [96]3 years ago

6 0

Problem would be easier to do if you'd write one function above the other:

<span>3x-2y=3; 6x-4y=1 should be:

</span><span>3x-2y=3
</span><span>6x-4y=1
------------

Please note that if you mult. the first equation by 2, you get 6x-4y= 6, which is almost but not quite the same as the second equation.
It should be obvious that 6x-4 in the first equation are the same expression as 6x-4 in the second equation. 6x-4 cannot = 6 and also =1 at the same time, so this system has NO solution. The 2 lines never intersect.

</span>

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cestrela7 [59]

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2 years ago

What value(s) for n will make this expression true?

exis [7]

Answer:

The answer is option 4.

Step-by-step explanation:

As there is a <em>M</em><em>o</em><em>d</em><em>u</em><em>l</em><em>u</em><em>s</em><em> </em><em>s</em><em>i</em><em>g</em><em>n</em><em> </em>so the answer will be always positive.

e.g

|1.2| = 1.2

|-5.6| = 5.6 (always interested in the positive value)

6 0

2 years ago

Read 2 more answers

What are the missing numbers?

Klio2033 [76]

Since it goes by thousandths (0.001), we can say the values are 2.089 and 2.095.

7 0

3 years ago

Read 2 more answers

Fill in the blank so that 81x2+_<br> x+81 is a perfect square trinomial.<br> 81x²+_____<br><br> x+81

densk [106]

the blank is 162

since 81 and 81 are both perfect squares the middle term (blank) will be the product of their roots times 2

√81=9

√81 =9

9 x 9 x 2

= 162

6 0

2 years ago

In a geometric sequence, a4 = 54 and a7 = 1,458. what is the 12th term? <br><br> answer: B) 354,294

slamgirl [31]

Option B:

The 12th term is 354294.

Solution:

Given data:

$a_4=54$ and $a_7=1458$

To find $a_{12}:$

The given sequence is a geometric sequence.

The general term of the geometric sequence is $a_n=a_1\ r^{n-1}$ .

If we have 2 terms of a geometric sequence $a_n$ and $a_k$ (n > K),

then we can write the general term as $a_n=a_k\ r^{n-k}$ .

Here we have $a_4=54$ and $a_7=1458$ .

So, n = 7 and k = 4 ( 7 > 4)

$a_7=a_4\ .\ r^{7-4}$

$1458=54\ . \ r^3$

This can be written as

$$r^3=\frac{1458}{54}$

$$r^3=27$

$$r^3=3^3$

Taking cube root on both sides of the equation, we get

r = 3

$a_{12}=a_7\ .\ r^{12-7}$

$=1458\ .\ r^5$

$=1458\ .\ 3^5$

$a_{12}=354294$

Hence the 12th term of the geometric sequence is 354294.

7 0

2 years ago