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2 years ago

Band Trip

Mathematics

1 answer:

Serhud [2]2 years ago

3 0

Answer:

The total costs for the trip, the hotel and the three one-day combination entrance is 224.(160+64=224)

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Z=Xm+n-p,solve for x

timama [110]

z=xm+n-p

We need to isolate x to one side I will do that on the left side so it looks cleaner, therefore I will first flip the equation.

xm+n-p=z

Now move the n and p to the right side

xm=z-n+p

x=(z-n+p)/m

Hope this helped, if you have any questions about this just comment and I will come answer.

6 0

3 years ago

5x minus 6 equal 44?

Lisa [10]

Answer:

x = 10

Step-by-step explanation:

I'm not entirely sure what you're asking, however we know that:

5x-6=44

5x must be 50, as 50-6=44

50 divided by 5 is 10, so 10x

6 0

3 years ago

Read 2 more answers

5/7 + 3/8?

Rudiy27

I don't know if anyone know say him answer

7 0

2 years ago

Keith started saving for retirement at age 45 with plans to retire at age 70. He invested an average of $500 per month in variou

saveliy_v [14]

500×((1+0.06÷12)^(12×25)−1)÷(0.06÷12)
=346,496.98

8 0

2 years ago

What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c

zysi [14]

Answer:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

Step-by-step explanation:

Given

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Required

Simplify

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}$

Cancel out 18

$\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}$

Divide 4 and 2

$\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}$

Divide 27 and 12 by 3

$\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}$

Apply law of indices

$\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}$

$\frac{9a^5 * b^3 * 2c }{4}$

Divide 2 and 4

$\frac{9a^5 * b^3 * c}{2}$

$\frac{9a^5b^3c}{2}$

Rewrite as:

$\frac{9}{2}a^5b^3c$

Hence:

$\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c$

4 0

2 years ago