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artcher [175]
3 years ago
8

What is 9,349 rounded to the nearest thousand?​

Mathematics
2 answers:
inn [45]3 years ago
7 0

Answer:

i think 9000

Step-by-step explanation:

zlopas [31]3 years ago
3 0

Answer:

9000

Step-by-step explanation:

since the hundreds value is under 5, you round down and it will round the thousands value to 9000

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Y=3x-8 in function notation
julia-pushkina [17]

To write it in function notation you'll convert it into a function, substituting y by f(x):


y = 3x - 8 = f(x) = 3x - 8



Hope it helped,



BioTeacher101

3 0
4 years ago
What is this answer?
Shtirlitz [24]

Answer:

1/2

Step-by-step explanation:

your mom :|

4 0
3 years ago
Read 2 more answers
Herbert has sold 99, 37, 86, and 73 copeirs in the last 4 months, respectivly. How many copiers will he need to sell this month
jolli1 [7]

Answer:

Therefore 80 ceperies will he need to sell this month.

Step-by-step explanation:

Average: Average is the ratio of sum of all numbers to the total number present in the data.

Given that Herbert has sold 99, 37, 86 and 73 copeirs in the last 4 months.

Let he need to sell x copeirs in this month.

According to the problem,

\frac{99+37+86+73+x}{5}=75

⇒ 99+37+86+73+x= 75×5

⇒295 + x= 375

⇒x = 375 - 295

⇒ x= 80

Therefore 80 ceperies will he need to sell this month.

3 0
3 years ago
Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

7 0
3 years ago
I need help. I don't understand ​
ycow [4]
Add them all together
5 0
3 years ago
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