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zhannawk [14.2K]

3 years ago

6

Drag each situation to show whether it can be modeled using a linear or an exponential function. If the situation cannot be mode

led with either function type, drag it into the container labeled Neither.

1 answer:

Goshia [24]3 years ago

5 0

Answer:

bihdbhdvn sfv jdn vhusrvnfg k,b snjbstdnj fcmranfbtgb,fy hje5rtgbv mdrhgsdmrmgbhedslrk cj5rtkolg segpo esjxmh9iorexdsvjbyhklycf,x n

Step-by-step explanation:

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A train leaves at 14:03 and arrives at 16:22. It travelled at an average speed of 95 km/h. How far did it travel? Give your answ

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Answer:

so it took it 2:03 to 4:22 it took the train an 2km to 50dp

Step-by-step explanation:

1 it is that you now 14 to 2 and 16 to 4 there you go

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3 years ago

Please help me find the function rule, i dont know what to do!! I will give big brain

Vesnalui [34]

The answer is Y=4x+5

6 0

2 years ago

Triangle ABC is an isosceles triangle. Angles B and C are base angles, with measurements of (11x −10)° and (7x + 10)°, respectiv

Ulleksa [173]

Answer:

m∠A = 90°

Step-by-step explanation:

In isosceles triangle base angles are congruent. That means we can equate measurments of angle B and angle C and solve for x!

m∠B = m∠C

11x - 10 = 7x + 10

4x - 10 = 10

4x = 20

x = 5

Now let's insert x back in the expressions for angles.

m∠B = (11x − 10)° = (11(5) − 10)° = 45°

m∠C = (7x + 10)° = (7(5) + 10)° = 45°

<u>Sum of all angles in the triangle is 180°.</u> Let's make an equation.

m∠A + m∠B + m∠C = 180°

m∠A + 45°+ 45° = 180°

m∠A = 90°

7 0

1 year ago

If h(x)= (f o g)(x) and h(x)= 4 √x+7, find g(x) if f(x) = 4 √x+1

stepladder [879]

(f o g)(x)=f(g(x))

so

h(x)=f(g(x))
$4\sqrt{x+7}=4\sqrt{g(x)+1}$
hmm, see the differences

x+7=g(x)+1

x+6=g(x)
g(x)=x+6

6 0

3 years ago

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

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