2 years ago

PLS HELP ASAP I WILL GIVE BRANIEST TO WHO ANSWERS THE QUESTION!!!!

Mathematics

1 answer:

marysya [2.9K]2 years ago

4 0

Answer:

10 miles is the correct answer

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What is the 50th term of the sequence that begins -4, 2, 8, 14....?

vagabundo [1.1K]

Answer:

$a_{50}=290$

Step-by-step explanation:

Notice that we are in the presence of an arithmetic sequence (there is a common difference of 6 between a term and the next). So thisis an arithmetic sequence with first term $a_1=-4$ . and common difference "<em>d</em>" = 6.

Recall now the formula for the nth term ( $a_n$ ) of an arithmetic sequence:

$a_n=a_1+(n-1)\,d$

Then, using our values in the formula, we get:

$a_n=a_1+(n-1)\,d\\a_{50}=-4+(50-1)\,6\\a_{50}=-4+(49)\,6\\a_{50}=290$

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3 years ago

Does this graph represent a function? Why or why not? Sey A. No, because it fails the vertical line test. B. Yes, because it has

Fittoniya [83]

Answer: it’s A..

Step-by-step explanation:

No, because it fails the vertical line test.

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3 years ago

Marshall began making a grilled cheese sandwich at 6:13 P.M. The sandwich was ready at 6:24 P.M. How long did it take Marshall t

jasenka [17]

It took him 11 minutes.

3 0

2 years ago

Read 2 more answers

One of the roots of the quadratic equation 4mnx^2 – 6m^2x – 6n^2x + 9mn = 0 (m, n ≠ 0) is a)-3m/2n b)3m/2n c)2m/3n d)-2m/3n

maxonik [38]

Answer:

$B.\ \frac{3m}{2n}$

Step-by-step explanation:

Given

$4mnx^2 - 6m^2x - 6n^2x + 9mn = 0\ (m, n \neq 0)$

Required

Calculate one of the root of the equation

$4mnx^2 - 6m^2x - 6n^2x + 9mn = 0$

Factorize

$2mx(2nx - 3m) -3n(2nx - 3m) = 0$

$(2mx - 3n)(2nx - 3m) = 0$

Split equation

$2mx - 3n = 0\ or\ 2nx - 3m = 0$

Make x the subject of formula in both expressions

$2mx = 3n\ or\ 2nx = 3m$

$x = \frac{3n}{2m}\ or\ x = \frac{3m}{2n}$

<em>From the list of given options, one of the roots of the equation is</em> $\frac{3m}{2n}$

4 0

3 years ago

PLEASE HELP ASAPPPPPPPP

Dvinal [7]

3.9 x 10 ^-5 is the answer. 0.000039 in scientific notation is 3.9 x 10^-5

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3 years ago

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