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3 years ago

Write an equation for the nth term of an arithmetic sequence then find a 25.

Mathematics

2 answers:

Damm [24]3 years ago

5 0

Look up math way and type it in its free and it will help alot.

I am Lyosha [343]3 years ago

3 0

19.

an<span> = 3n + 2

...........................

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1 quart is equal to 4 cups given this fact what is 8 quarts equivalent to in cups

SVETLANKA909090 [29]

8 quarts is equal to 32 cups :)

5 0

3 years ago

One large bubble separates into four small bubbles so that the total area of the small bubbles is equal to the area of the large

anzhelika [568]

Answer:

10 cm.

Step-by-step explanation:

We'll begin by calculating the area of the small bubble. This can be obtained as follow:

Radius (r) = 5 cm

Area (A) =?

Since the bubble is circular in nature, we shall use the formula for area of circle to determine the area of the bubble. This is illustrated below:

A = πr²

A = π × 5²

A = 25π cm²

Next, we shall determine the total area of the small bubbles. This can be obtained as follow:

Area of 1 bubble = 25π cm²

Therefore,

Area of 4 bubbles = 4 × 25π cm²

Area of 4 bubbles = 100π cm²

Finally, we shall determine the radius of the large bubble. This can be obtained as follow:

Area of large bubble = total area of small bubbles = 100π cm²

Radius (r) =?

A = πr²

100π = πr²

100 = r²

Take the square root of both side

r = √100

r = 10 cm

Thus, the radius of the large bubble is 10 cm

5 0

2 years ago

Can someone help me please

Lisa [10]

Answer:

perimeter is 12x-6

Step-by-step explanation:

perimeter is the sum of all sides

3(x+2) + 2x+11 + 7x-23

distribute the 3 in the first term, then combine 'like terms'

3x+6 + 2x+11 + 7x-23

3x+2x+7x + 6+11+(-23)

12x-6

4 0

1 year ago

Read 2 more answers

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$