All categories

2 years ago

Solve the inequality. Graph the solution set. 23 + 156 > 5(3b + 1)

Mathematics

1 answer:

Elan Coil [88]2 years ago

4 0

Answer:

b<58/5

Step-by-step explanation:

You might be interested in

Mrs. Kang gave Caleb a collection of 25 marbles. Each month Caleb adds 4 marbles to his collection. Which inequality can be used

GarryVolchara [31]

Answer:

I did the math and got an answer of 17.5

Step-by-step explanation:

I divided 70 by 4 and got a result of 17.5. This probably means Caleb started his collection about 17 and 1/2 months ago. (I apologize if it's wrong)

3 0

2 years ago

Read 2 more answers

I need these answers for the photos

Jobisdone [24]

That would be -2 1/2 , 3 1/2

4 0

2 years ago

How to solve x/3 + 4 = -2

r-ruslan [8.4K]

Answer:

x=−18

Step-by-step explanation:

Let's solve your equation step-by-step.

+4=−2

Step 1: Simplify both sides of the equation.

x+4=−2

Step 2: Subtract 4 from both sides.

x+4−4=−2−4

x=−6

Step 3: Multiply both sides by 3.

3*(

x)=(3)*(−6)

x=−18

5 0

2 years ago

Read 2 more answers

This is for my sister she need help ASAP

lisov135 [29]

Answer:

Your Answer is C

3 0

2 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

2 years ago