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3 years ago

I am literally so sorry but I DON'T UNDERSTAND THIS IN MY MATH CLASS! PLEASE HELP

Mathematics

2 answers:

STALIN [3.7K]3 years ago

7 0

Answer:

− 1.9 f − 25

Step-by-step explanation:

You have to simplify the terms -2.3f + 0.4f and 16 - 9

hope this helped :))

zaharov [31]3 years ago

6 0

I think the answer is -2.7f-25 because if you subtract -2.3 F + .4 F then you do -16-9 I could be wrong warning

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In Art Class, you designed a rectangle stained glass window. The

Ierofanga [76]

Answer:

Step-by-step explanation:

Draw a triangle and label the sides (picture below):

● the length is 2x+2

● and the width or the height is 6x

The perimeter of a rectangle is the sum of the sides

● P = 2x+2 + 6x + 2x+2 + 6x

Isolate the similar terms

● P = (2x+2x+6x+6x) + (2+2)

● P = 16x + 4

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3 years ago

Can someone help me out with this problem

Nostrana [21]

I need more information.

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2 years ago

Suppose y = 48 + 3(2n - 1) is an explicit representation of an arithmetic sequence for integer values n ≥ 1. Find the xth partia

omeli [17]

$a_n=48+3(2n-1)$

The formula of the sum of the arithmetic sequence:
$S_n=\dfrac{a_1+a__n}{2}\cdot n$
calculate:
$a_1=48+3(2\cdot1-1)=48+3=51$
substitute
$S_n=\dfrac{51+48+3(2n-1)}{2}\cdot n=\dfrac{99+6n-3}{2}\cdot n=\dfrac{96+6n}{2}\cdot n=3n^2+48n$
Your answer is:
$\boxed{f(x)=3x^2+48x}$

5 0

3 years ago

Read 2 more answers

Line p passes through A(-2, -4) and has a slope of 1/2 What is the standard form of the equation for line p?

yarga [219]

Point slope form is y-(-4) = 1/2(x-(-2)

y+4 =1/2 (x+2)

now the standard form 2y+8=x+2
so -x+2y=-6 is the equation for line p in standard form

4 0

3 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

3 years ago