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Goryan [66]
1 year ago
15

T+b = 41 t = 2b+2 Solve for t and b

Mathematics
1 answer:
erastovalidia [21]1 year ago
3 0

Answer: b=13, t=29

Step-by-step explanation:

Substituting the first equation into the second,

2b+2+b=41\\\\3b+2=41\\\\3b=39\\\\b=13\\\\\implies t=2(13)+2=29

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Gemiola [76]

Answer:

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Step-by-step explanation:

3 0
3 years ago
What is equivalent to12/18
solmaris [256]
  <span>2/3 and 24/36 and 6/9 
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3 years ago
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The volume of a large aquarium is 210 yd. It is 3 yd wide and 2
Pavel [41]

Length of the aquarium is 35 yd

Step-by-step explanation:

  • Step 1: Volume of the aquarium = 210 yd³ given by Volume = length × width × height. Width = 3 yd and Height = 2 yd

Length = Volume/Width × Height

            = 210/3 × 2 = 210/6

            = 35 yd

5 0
2 years ago
A publisher reports that 54% of their readers own a particular make of car. A marketing executive wants to test the claim that t
dalvyx [7]

Answer and explanation:

Null hypothesis: p is not 54%

Alternative hypothesis : p is equal 54%

First find standard deviation

Standard deviation is:

σ = √[ P ( 1 - P ) / n ] = √[ 0.54×0.46/220 ] =0.033

The z score is calculated p-P/σ = (0.54-0.50)/0.033 = 1.2121

Therefore we can conclude that proportions are not same and fail to reject the null hypothesis

3 0
3 years ago
Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

f'(x)=2(3)x^{2}+3(2)x^1-336

f'(x)=6x^{2}+6x-336

6x^{2}+6x-336=0

x^{2}+x-56=0

x^{2}+8x-7x-56=0

(x+8)-7(x+8)=0

(x+8)(x-7)=0

x=-8,x=7

So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

Now we will find derivative over each interval as:

f'(x)=(x+8)(x-7)

f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16

Since f'(9)>0, therefore, function is increasing on interval (-\infty,-8).

f'(x)=(x+8)(x-7)

f'(1)=(1+8)(1-7)=(9)(-6)=-54

Since f'(1), therefore, function is decreasing on interval (-8,7).

Let us check for the derivative at x=8.

f'(x)=(x+8)(x-7)

f'(8)=(8+8)(8-7)=(16)(1)=16

Since f'(8)>0, therefore, function is increasing on interval (7,\infty).

(b) Since x=-8,7 are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

f(-8)=2(-8)^3+3(-8)^2-336(-8)

f(-8)=1856

f(7)=2(7)^3+3(7)^2-336(7)

f(7)=-1519

Therefore, (-8,1856) is a local maximum and (7,-1519) is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

f''(x)=12x+6

12x+6=0

12x=-6

\frac{12x}{12}=-\frac{6}{12}

x=-\frac{1}{2}

Therefore, x=-\frac{1}{2} is an inflection point of given function.

3 0
2 years ago
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