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2 years ago

Can someone solve 8 = b - 1 + 8b STEP BY STEP PLEASE

Mathematics

1 answer:

andrew11 [14]2 years ago

7 0

Answer:

8 = b - 1 + 8b

group like terms

8 + 1 = b + 8b

9 = 9b cus b standing alone me 1b

divide both side by 9

9 = 9b

9 9

b = 1

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Write your question here (Keep it simple and clear to get the best answer)what number is 20times table

kykrilka [37]

20×1=20
20×2=40
20×3=60
20×4=80
20×5=100
20×6=120
20×7=140
20×8=160
20×9=180
20×10=200
(Only include this if the table is up to 12)
20×11=220
20×12=240

6 0

3 years ago

Simplify 30 – 5(15 - 10). A. -55B. 125C.20D.5

Harrizon [31]

Simplify : 30 – 5(15 - 10).

Apply BODMAS :

B- Brackets

O-Of

D-Division

M-Multiplication

A-Adition

S-Subtraction

$\begin{gathered} 30-5(15-10) \\ 30-5(5) \\ 30-25 \\ 5 \end{gathered}$

Answer : D) 5

6 0

8 months ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

If the area of a circle is 50.25 meters what is the radius

NARA [144]

Answer:

≈3.99938

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Y=x+2 x-y=2 What is the solution

Alisiya [41]

Answer:

No solutions.

Step-by-step explanation:

x - y = 2 <Start with an equation

x - (x + 2) = 2 <Plug in the given y value and solve for x

x - x - 2 = 2 <Combine like terms

0 - 2 = 2 <There are no values of x

No solutions.

8 0

3 years ago