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1 year ago

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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Answer:

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3 years ago

What is equivalent to12/18

solmaris [256]

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3 years ago

Read 2 more answers

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

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3 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

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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.

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

3 years ago