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2 years ago

the value of the nth term in a sequence of numbers is given by the expression 2n-3. What is the 8th term in the sequence

Mathematics

2 answers:

Debora [2.8K]2 years ago

8 0

Answer:

2n-3=0

2n=3

n=3/2

therefore answer is 3/2

ValentinkaMS [17]2 years ago

8 0

Answer:

bro I didn't even look up this question and it showed up so I don't know dude

Step-by-step explanation:

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Find the slope of the line that passes through (-11, 75) and (27, -9).

Marrrta [24]

Answer:

$\frac{y__2-y__1}{x__2-x__1}$

y = -9 + 75 = 66

x = 27-(-11) = 27 + 11 = 38

slope = 66/28

5 0

2 years ago

The sum of 4 and the product of 5 and x is equal to 5 less than the product of 2 and x. What is the value of x?

Alexxandr [17]

Answer:

b. -3

Step-by-step explanation:

i hope this helps :)

7 0

2 years ago

PLEASE HELP!! i dont understand it

andreev551 [17]

Answer:

a. y^ -1 = e^x +2

Step-by-step explanation:

y = ln (x-2)

Exchange x and y

x = ln (y-2)

Solve for y

Raise each side with a base of e

e^ x = e^(ln(y-2)

e^x = y-2

Add 2 to each side

e^x +2 =y

6 0

2 years ago

Answer all & show work! WILL MARK BRAINLIEST!

Eddi Din [679]

Answer:

Step-by-step explanation:

1 12

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6 0

2 years ago

Suppose that the functions s and t are defined for all real numbers x as follows.

jeka94

Answer:

$(s.t)(x) = 4x^2+24x^2\\(s-t)(x) = x+6-4x^2\\(s+t)(-3) = 39$

Step-by-step explanation:

Given functions are:

$s(x)= x+6\\t(x)= 4x^2$

We have to find:

(s.t)(x) => this means we have to multiply the two functions to get the result.

So,

$(s.t)(x) = s(x)*t(x)\\= (x+6)(4x^2)\\=4x^2.x+4x^2.6\\=4x^3+24x^2$

Also we have to find

(s-t)(x) => we have to subtract function t from function s

$(s-t)(x) = s(x) - t(x)\\= (x+6) - (4x^2)\\=x+6-4x^2$

Also we have to find,

(s+t)(-3) => first we have to find sum of both functions and then put -3 in place of x

$(s+t)(x) = s(x)+t(x)\\= x+6+4x^2$

Putting x = -3

$= -3+6+4(-3)^2\\=-3+6+4(9)\\=3+36\\=39$

Hence,

$(s.t)(x) = 4x^2+24x^2\\(s-t)(x) = x+6-4x^2\\(s+t)(-3) = 39$

8 0

2 years ago