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2 years ago

Help photo question. Mathematics. :D

Mathematics

1 answer:

Aleks04 [339]2 years ago

5 0

M = 4
n = 1

If you substitute in your values, you can see that 4 + 1 = 5
And 4 - 1 = 3
Hope this helped :)

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The formula for the circumference of a circle is C = 2nr. Find the circumference of a

Sholpan [36]

Step-by-step explanation:

$circumference = 2\pi \: r \\ C = 2 \times 3.14 \times ( \frac{4}{2} ) \\ C = 12.56 \: ft$

7 0

2 years ago

Given that A varies directly as B and inversely as C and that; A=12 when B=3 and C=2. Find B when A=10 and C=1.5

maria [59]

Answer:

B = 1.875

Step-by-step explanation:

given that A varies directly as B and inversely as C then the equation relating them is

A = $\frac{kB}{C}$ ← k is the constant of variation

to find k use the condition A = 12 when B = 3 and C = 2 , then

12 = $\frac{3k}{2}$ ( multiply both sides by 2 to clear the fraction )

24 = 3k ( divide both sides by 3 )

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8 0

1 year ago

What would be the answer to this question?

Iteru [2.4K]

The equation has one solution, x = 0
3x = 4x
-3x both sides
0 = 4x - 3x
0 = x

3 0

2 years ago

In order to explain finicial concepts to your clients you need to have

Neporo4naja [7]

Im think its C. Accounting.

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7 0

2 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$

6 0

2 years ago