All categories

3 years ago

What is answer for b the variable

Mathematics

1 answer:

borishaifa [10]3 years ago

6 0

Salutations!

$-35 = 7b$

Isolate the number

$-35/7 = b$

$-5 = b$

$b$ is equal to $-5$

Hope I helped (:

Have a great day!

You might be interested in

Which statement is false ? 1. Every integer is also a real number , 2. No irrational number is rational , 3. Every integer is al

Whitepunk [10]

Every integer is also an irrational number...false

-2 is an integer...and it is rational because it can be made into a fraction..-2/1

3 0

4 years ago

A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the person's weight, in pounds, when rounded to th

Vlada [557]

Answer:

you multiply 0.45kg by 87kg you will get 39lb

8 0

3 years ago

(due by 9)

ValentinkaMS [17]

Answer:

n=6 1/3

Step-by-step explanation:

add -4/3 to both sides

5 0

3 years ago

Read 2 more answers

Faces of a three-dimensional figure are any of the __________________, including the bases.

tatiyna

Answer:

A prism is a three-dimensional figure with:

o at least two parallel, congruent faces called bases that are polygons.

o Named by its base

 Examples: triangular prism, rectangular prism

 A pyramid is a three-dimensional figure with:

o one base that is a polygon.

o Its other faces are triangles.

o Examples: triangular pyramid, square pyramid, rectangular pyramid

Step-by-step explanation:

4 0

3 years ago

Derive the first three (non-zero) terms of Taylor's series expansion for the function

mestny [16]

Answer:

We want to find the first 3 terms of the Taylor's series expansion for f(x) = sin(x) around x = 0.

Remember that a Taylor's series expansion of a function f(x) around the point x₀ is given by:

$f(x) = f(x_0) + \frac{1}{2!}f'(x_0)*(x - x_0) + \frac{1}{3!}*f''(x_0)*(x - x0)^2 + ...$

Where in the formula we have the first 3 terms of the expansion (but there are a lot more).

So, if:

f(x) = sin(x)

x₀ = 0

The terms are:

$f(x_0) = sin(0) = 0$

$\frac{1}{2!}*f(x_0)'*(x - x_0) = \frac{1}{2} cos(0)*(x - 0) = x/2$

$\frac{1}{3!}*f(x_0)''*(x - x_0)^2 = \frac{1}{6}*-sin(0)*(x - 0)^2 = 0$

$\frac{1}{4!}*f(x_0)'''*(x - x_0)^3 = \frac{1}{24}*-cos(0)*(x- 0)^3 = -\frac{x^3}{24}$

We already can see that the next term is zero (because when we derive the cos part, we will get a sin() that is zero when evaluated in x = 0), then the next non zero term is:

$\frac{1}{6!}*f(x_0)''''*(x - x_0)^5 = \frac{1}{2*3*4*5*6} *(x - 0)^5 = \frac{x^5}{720}$

Then we can write:

$sin(x) = \frac{x}{2} - \frac{x^3}{24} + \frac{x^5}{720}$

Evaluating this in x = 0.2, we get:

$sin(0.2) = \frac{0.2}{2} - \frac{0.2^3}{24} + \frac{0.2^5}{720} = 0.099667$

7 0

3 years ago