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worty [1.4K]
3 years ago
6

Which transformations would affect the starting point of a square root function?

Mathematics
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

There are at least two transformations that would affect the starting point of a square root: (i) <em>Horizontal translation</em>, (ii) <em>Vertical translation</em>.

Step-by-step explanation:

Let the square be represented by the following function:

f(x) = \sqrt{x},\,\forall \,x \ge 0 (1)

Which means that function has only valid solutions for every element of x greater or equal to 0. In particular, the starting point is x = 0.

We can change the starting point of this function by horizontal translation, which is defined below:

g(x) = f(x - a), \,\forall \,x \ge a (2)

In other words, we create the resulting function:

g(x) = \sqrt{x-a}, \,\forall\,x \ge a (3)

Another, possibility is using a vertical translation, whose definition is defined below:

h(x) = f(x)+c,\,\forall \,x \ge 0 (4)

In other words, we create the following function:

h(x) = \sqrt{x} +c (5)

Hence, there are at least two transformations that would affect the starting point of a square root: (i) <em>Horizontal translation</em>, (ii) <em>Vertical translation</em>.

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Enter the correct answer in the box. What is this expression in simplified form? <br> 3\sqrt{88}
Ugo [173]

Answer: 12\sqrt{5}

Step-by-step explanation:

3\sqrt{88}

88 can be rewritten as 8 x 10 and 8 can be rewritten as 2^2*2 making a 2 come out of the root.

3\sqrt{2^2*2*10}\\ 3*2\sqrt{2*10}\\6\sqrt{2*10}

10 can be rewritten as 2 x 5.

6\sqrt{2*2*5}

2*2 can also be rewritten as 2^2 making a 2 come out of the root.

6\sqrt{2^2*5}\\ 6*2\sqrt{5}\\ 12\sqrt{5}

We can't keep simplifying the root.

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3 years ago
Austin will rent a car for the weekend. He can choose one of two plans. The first plan has an initial fee of $40 and costs an ad
GenaCL600 [577]

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Step-by-step explanation:

7 0
1 year ago
Plsss help no one answered the other times I put this question
svetlana [45]

Answer:

a) <em>The equation</em> (10s + 8w) <em>represents </em><em>the </em><em>calories </em><em>Bridget </em><em>ate </em><em>on </em><em>Monday </em><em>and </em><em>the </em><em>equation</em> (20s + w) <em>represents </em><em>the </em><em>calories</em><em> </em><em>she </em><em>ate</em><em> </em><em>the </em><em>next </em><em>day.</em>

<em>b)</em><em> </em><em>The </em><em>number </em><em>of </em><em>calories</em><em> </em><em>in </em><em>each </em><em>strawberry</em><em> </em><em>is </em>4 <em>and </em><em>the </em><em>number </em><em>of </em><em>calories </em><em>in </em><em>each </em><em>vanilla</em><em> </em><em>wafer</em><em> cookie</em><em> </em><em>is </em>19. The solution is s= 4 and w = 19.

Step-by-step explanation:

For part A, Bridget ate 10 strawberries and 8 vanilla wafer cookies on Monday. Since the the number of calories in a strawberry is <em>s</em> and the number of calories in a vanilla wafer cookie is <em>w </em>, the number of calories Bridget ate on Monday is <em>10s + 8w</em><em>.</em><em> </em>The next day, Bridget ate 20 strawberries and 1 vanilla wafer cookie. Hence, the number of calories Bridget ate on the next day is 20s<em> + w</em>.

For part B,

we will create two different simultaneous equations.

Equation 1: 10s + 8w = 192

Equation 2: 20s + w = 99

We need to find one of the terms first to solve the other term. For this case, I will solve for w first.

Multiply the first equation by 2.

Equation 3: 20s + 16w = 192*2 = 384.

Now, subtract equation 2 from this new equation.

Equation 4:

(20s + 16w) - (20s + w) = 384 - 99

20s + 16w - 20s - w = 285

15w = 285

This leaves only w left and we can solve w.

w = 285 / 15 = 19

Now, we can solve for s using equation 2.

20s + 19 = 99

20s = 99-19 = 80

Hence,

s = 80/20 = 4

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Answer:

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Step-by-step explanation:

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