2 years ago

The difference of two numbers is nine the larger number is one more than find the larger number

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

6 0

Answer:

Area = 36π mm²

Step-by-step explanation:

The first question is incomplete

Below is the solution to attached question

Area of circle = πR²

Where R = radius = 6mm

Area of circle = π× 6²

Area = π×36

Area = 36π mm²

I hope this was helpful, please mark as brainliest

You might be interested in

HELP ASAP

stira [4]

Answer:

Use khan academy for the graph.

Step-by-step explanation:

8 0

2 years ago

BRAINLIEST for whoever gets this correct!

ratelena [41]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find the constant of proportionality in the graph

pshichka [43]

The answer is B. Since the graph increases by 4 on the y-axis as <em>x</em> increases by 1, the constant of proportionality is 4.

7 0

2 years ago

PLEASE HELP PICTURE SHOWN CHECK ALL THAT APPLY

marusya05 [52]

To find the zeros set the factors on the numerator equal to zero and solve
(x - 1) = 0 x = 1
(x + 1) = 0 x = -1

5 0

2 years ago

Rewrite 2tan3x in terms of tanx

blsea [12.9K]

$2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}$

Use the same identity to expand $\tan2x$ .

$\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}$

$\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}$
$2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}$
$2\tan3x=\dfrac{2\tan x\left(2+1-\tan^2x\right)}{1-\tan^2x-2\tan^2x}$
$2\tan3x=\dfrac{2\tan x(3-\tan^2x)}{1-3\tan^2x}$

4 0

3 years ago