All categories

3 years ago

On a number line 8.2 would be located where

Mathematics

1 answer:

lesya [120]3 years ago

7 0

Answer:

Step-by-step explanation:

8.2 comes between 8 and 9

8.2 = $8\frac{2}{10}$ .

As you know, the interval between 8 and 9 is divided into 10 equal parts and each part is 1/10 or 0.1.

You might be interested in

An inequality can have more than one solution

ratelena [41]

True. If there is a point in the middle that is false, we have two create 2 solutions.

Hope this helps!

4 0

3 years ago

Can you solve the inequality 2(x-3) is less than or equal to 10 without using the distributive property? Explain.

STatiana [176]

Answer:

Yes, without using the distributive property, we get the value for the given expression 2(x - 3) ≤ 10 as x ≤ 8

Step-by-step explanation:

Here, the given expression is:

2(x - 3) ≤ 10

Yes, the given expression can be solved without using the DISTRIBUTIVE Property

Here, consider the given expression:

2(x - 3) ≤ 10

Now, divide the inequality by 2 on both sides, we get:

$\frac{2(x-3)}{2} \leq \frac{10}{2}\\\implies (x - 3) \leq 5$

Now adding (+3) on both the sides, we get:

(x - 3) ≤ 5 ⇒ (x - 3 + 3 ) ≤ 5 + 3

or, x ≤ 8

Hence, without using the distributive property, we get the value of x ≤ 8 for the given expression 2(x - 3) ≤ 10

6 0

3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago

Round 231469.335329 to the nearest thousand

lorasvet [3.4K]

Answer: 231,469.335

Step-by-step explanation:

rounded to the nearest 0.001 or the thousandths place

7 0

3 years ago

Read 2 more answers

I need this answered

Ilya [14]

Hope this makes sense! It was too hard to type out lol. The one on the left is how it is solved the and one on the right kinda explains how I got the equation.

8 0

2 years ago