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2 years ago

The prime

Mathematics

1 answer:

Serjik [45]2 years ago

8 0

9514 1404 393

Answer:

n = 3

Step-by-step explanation:

In order for 3 × 3 = 9 to be a factor of both numbers, we must have ...

n = 3

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Find the perimeter of the figure?

zhannawk [14.2K]

Answer:

Step-by-step explanation:

9+9+9+9=36

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3 years ago

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Suppose you invest $16000 at 9% interest and that it is compounded daily. How much will you have in 8 years?

kramer

Answer:

The amount after 8 years is $ 16,031.579

Step-by-step explanation:

Given as :

The Principal invested = $ 16000

The rate of interest compounded daily = 9 %

The time period = 8 years

Let The amount after 8 years = $ A

<u>From Compounded method </u>

Amount = Principal invested × $(1+\dfrac{\textrm Rate}{365\times 100})^{365\times \textrm Time}$

Or, Amount = 16000 × $(1+\dfrac{\textrm 9}{365\times 100})^{365\times \textrm 8}$

Or, Amount = 16000 × $(1.0002465)^{8}$

∴ Amount = $ 16,031.579

Hence The amount after 8 years is $ 16,031.579 Answer

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3 years ago

8640 divided by40 long division

Tju [1.3M]

The answer is 216.
https://photomath.net/s/YLLqQX

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3 years ago

Find the lateral area and total area of a regular pyramid with a square base. The bae is 10m and the slant edge or height is 13m

kogti [31]

the lateral area and total area of regular pyramid with a square base is 130

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2 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago