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2 years ago

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The linear functions f(x) and g(x) are represented on the graph, where g(x) is a transformation of f(x): A graph with two linear

functions; f of x passes through 1, 3 and 3, 13, and g of x passes through negative 1, 3 and 1, 13.

1 answer:

Valentin [98]2 years ago

8 0

Answer:

to be fr idk sryy

Step-by-step explanation:

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Match each operation involving f(x) and g(x) to its answer.

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F(x)=1-x² and g(x)=√(11-4x)

(g+f)(2)=>

1-(2)²+√(11-4(2))

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(g-f)(-1)

√(11-4*-1)-(1-(-1)²

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-3√3

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2 years ago

A dealer has a lot that can hold 30 vehicles. In this lot, there are two available models, A and B. The dealer normally sells at

Vanyuwa [196]

The answer is a 20 model A, 10 model B

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2 years ago

Write the first five terms of the geometric sequence in which a1=64 and the common ratio is 5/4

natita [175]

Remember that finding terms in a geometric sequence is done by multiplying the previous term by a common ratio $r$ . For example, we can say:

$a_2 = a_1 r$

$a_3 = a_2 r = (a_1 r)r = a_1 r^2$

We have $a_1 = 64$ . To find $a_2$ , let's multiply this term by $\frac{5}{4}$ :

$a_2 = 64 \cdot \frac{5}{4} = 80$

Now, let's use this to find all of our other terms:

$a_3 = 80 \cdot \frac{5}{4} = 100$

$a_4 = 100 \cdot \frac{5}{4} = 125$

$a_5 = 125 \cdot \frac{5}{4} = \frac{625}{4}$

Thus, our terms are 64, 80, 100, 125, and (625/4).

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2 years ago

PLEASE HELP: For the next RSM Award ceremony, it was decided to decrease the length of the main poster by 12% and to increase i

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Let me set an example to help solve this, say the poster is 100 inches by 90 inches. the length of the poster is reduced by 12% so subtract 12 from 100 to get you new length of 78 inches. the width is increased by 15% so multiply 90 by 15% to get 13.5 inches and add the to the 90 inches you have to get your new width of 103.5 inches. multiply them together to get your new area then divide the new area by the old area, subtract that number by 1 and multiply it by -1.

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2 years ago

Oil leaks out of a tanker at a rate of r=f(t) liters per minute, where t is in minutes. If f(t)=Ae^(−kt), write a definite integ

Slav-nsk [51]

Answer:

The definite integral expressing the total quantity of oil, 'V', which leaks out of the tanker in the first hour is given as follows;

$V = \int\limits^{60}_0 {A \cdot e^{(-k \cdot t)}} \, dt$

Step-by-step explanation:

From the question, we have;

The rate at which oil leaks out of the tanker, r = f(t)

The unit of the oil leak = Liters per minute

The unit of t = Minutes

$If \ f(t) = A \cdot e^{(-k \cdot t )}$

Therefore, we have;

The definite integral expressing the total quantity, 'V', of oil which leaks out of the tanker in the first hour is given as follows;

$V = \int\limits^{60}_0 {A \cdot e^{(-k \cdot t)}} \, dt$

Therefore, we have;

$\int\limits^{60}_0 {A \cdot e^{(-k \cdot t)}} \, dt = \dfrac{A \cdot e ^{60 \cdot k} - A}{k \cdot e ^{60 \cdot k} }$

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2 years ago