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timurjin [86]
2 years ago
11

Lance baked a pie Justin ate 2/3 of the pie and Joey ate 1/4 of the pie how much of the pie is left

Mathematics
1 answer:
zepelin [54]2 years ago
8 0
1 - 2/3 - 1/4
= 1 - 8/12 - 3/12
= 1/12
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#4 find the value of x round answer to the nearest tenth
AnnyKZ [126]

Answer:

Step-by-step explanation:

Since this is a right angled triangle:

tan(∅) = opposite side/adjacent side

In this triangle:

tan(25°) = x/19

x = 8. 86 ≅ 8.9

6 0
3 years ago
Read 2 more answers
Order of Operations- -6+(-30) divided by 5-(-2). Explain step by step
Reil [10]

Answer:

-36/7 or -5.14287

Step-by-step explanation:

so I would get rid of parentheses you do not need them

\frac{-6+-30}{5--2}=\frac{-36}{7}(because when two negative come before two numbers like -x-y they are being added but negative is present to the result of the sum, when one negative sign comes after a number and the other negative sign comes before a number, like x--y they are being added as well but negative is not present to the result of the sum)

-36/7=-5.1428

hope this helps!

7 0
3 years ago
Can someone please help me??
Nesterboy [21]

Answer:

Answers are below

Step-by-step explanation:

x     y

0    $25,001.00

1     $25,000.95

2    $25,000.90

3    $25,000.86

I graphed the function on the graph below and found the values of y for each value of x.

*My graph showed that the initial value of the printer was $25,001.00, instead of $25,000.00*

7 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
The function h(t) = –16t^2 + 32t + 48 is graphed. When will h(t) = 48?
lana [24]

Answer:

h(0) = –16t^2 + 32t + 48

Step-by-step explanation:

If everything with a t is multiplied by zero, it will make everything 0 and leave the value of the constant, 48.

3 0
3 years ago
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