: figure below, find the exact value of z.

I need help with this problem. <br>

Zina [86]

Answer:

(1,4)

Step-by-step explanation:

Let's this of $y=\sqrt{x}$ which is it's parent function.

How do we get to $y=-\sqrt{x-1}+4$ from there?

It has been reflected about the a-axis because of the - in front of the square root.

It has been shifted right 1 unit because of the -(1) in the square root.

It has been moved up 4 units because of the +4 outside the square root.

In general:

$y=a(x-h)^2+k$ has the following transformations from the parent:

Moved right h units if h is positive.

Moved left h units if h is negative.

Moved up k units if k is positive.

Moved down k units if k is negative.

If $a$ is positive, it has not been reflected.

If $a$ is negative, it has been reflected about the x-axis.

$a$ also tells us about the stretching factor.

The parent function has a starting point at (0,0). Where does this point move on the new graph?

It new graphed was the parent function but reflected over x-axis and shifted right 1 unit and moved up 4 units.

So the new starting point is (0+1,0+4)=(1,4).

6 0

3 years ago

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Choose the pair of numbers that is in the relationship shown

ozzi

Answer:

First one is the correct answer!!

Step-by-step explanation:

6 0

2 years ago

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

3 years ago

How many pairs of angles are marked as congruent in the figures below?

MrRissso [65]

Answer:2

Step-by-step explanation:According to the diagram,there is a single angle marking for one pair of angles V and M,a double angle marking for angles X and L.

5 0

3 years ago

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A cheese pizza costs $5. Additional toppings cost $1.50 each. Write and graph an equation that represents the total cost $c$ (in

Triss [41]

Answer:

5 + 1.5x = total cost

Step-by-step explanation:

5 is for one pizza and x is for toppings. For every topping it is multiplied by 1.5. for example, if you have 2 toppings for one cheese pizza it would be 5 dollars + 1.5x2 which is 3 so 3+5=8$ in total.

7 0

3 years ago

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