A recipe call for 2 eggs to make 10 pancakes. How many eggs will you need to make 35 pancakes?

Please answer the following question. Also please answer if you actually know it.

Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated 4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8

5 0

3 years ago

The center of a circle and a point on the circle are given. Write the equation of the circle in standard form.

pogonyaev

Answer:

A.Because ,center is (3,2)and point is (-2-3) .

and answer is (x-3)2+(y+2)2=52.

6 0

3 years ago

Which function describes the arithmetic sequence shown -1, 1, 3, 5, 7, 9, 11, 13, ...

riadik2000 [5.3K]

Answer:

f(n) = 2n - 3 describes the arithmetic sequence,

-1, 1, 3, 5, 7, 9, 11, 13,.......

Step-by-step explanation:

Function:

A function is like a machine that gives an output for a given input.

A function has an independent variable which is called the input of the function.

The output for a given input is called the dependent variable.

Here. 'n' is the independent variable

f(n) is the function dependent variable i.e function of n as an output.

For Arithmetic sequence

$a_{n}= a_{1} + (n-1)d$

Where

$a_{n}$ = nth number of the sequence.

a₁ = First term

d = common difference = a₂ - a₁ = a₃ - a₂ = so on

For a function which describes Arithmetic sequence,

a₁ = -1

d = 1 - -1 =2

∴ f(n) = a₁ + (n -1 )d

substituting the values we get

∴ f(n) = -1 +(n-1)2

using distributive property

∴ f(n) = - 1 + 2n - 2

∴ f(n) = 2n - 3 ..........is the required function

7 0

3 years ago

A school district wants to determine the most effective way to prepare seniors for the SAT. They randomly select senior level ma

My name is Ann [436]

Answer:

In the SAT; Pedagogical relationships are not traditional but those of a group of people working together towards a common goal. Students and tutor challenge specific problems together and face a style of encounter that opens the way to self-learning.

● Within the SAT group, which is at the same time a group of neighbors who share possibilities and problems, learning to dialogue is both an educational objective and a means of training.

Step-by-step explanation:

The purpose of the SAT is for graduates to be competent to perform in their own communities at different levels of activity. The Booster, for example, works at the level of improvement of productive projects within its own productive units or those of its neighbors. The Pilot, in addition to the previous capacity, can organize different types of activities for the common good (health projects, the environment, work with children, recreation, production in solidarity). These two levels are approved by law with the basic secondary. At the Bachelor level, graduates are trained to participate in more complex organizations and develop long-term initiatives in both productive and organizational aspects.

To achieve greater ease and adaptation to the SAT, I consider that the group that is being taught with the non-traditional method and where the students work in groups, can easily adapt to the SAT learning system.

4 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago