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3 years ago

Vertex form is f(x)=a(x-p)^2 +q. How do i determine a?

Mathematics

1 answer:

slamgirl [31]3 years ago

8 0

Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.

y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).

Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,

-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.

The equation of parabola y4 is y+4 = (1/4)x^2

Or you could elim. the fraction and write the eqn as 4y+16=x^2, or

4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.

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Moby sold half his toy car collection, and then bought 12 more cars the next day. He now has 48 toy cars. How many cars did he h

castortr0y [4]

Answer: 72 cars.

Step-by-step explanation:

Let be "x" the number of toys that Moby had to start with.

According to the information given in the exercise, Moby sold half his toy car collection. This can be represented with the following expression:

$\frac{1}{2}x$

Then he bought 12 more cars, giving a total amount of 48 toy cars.

Therefore, the equation that represents this situation is the equation shown below:

$x-\frac{1}{2}x+12=48$

Now you must solve for "x" in order to find its value.

You get that this is:

$\frac{1}{2}x+12=48\\\\\frac{1}{2}x=48-12\\\\\frac{1}{2}x=36\\\\x=(36)(2)\\\\x=72$

7 0

4 years ago

What is the length of segment AE?

nexus9112 [7]

Answer:

$AE=\frac{50}{3}\ units$

Step-by-step explanation:

step 1

In the right triangle ABC

Applying the Pythagoras Theorem fin the hypotenuse AC

$AC^{2} =AB^{2}+BC^{2}$

substitute

$AC^{2} =6^{2}+8^{2}$

$AC^{2} =100$

$AC =10\ units$

step 2

we know that

If two figures are similar, then the ratio of its corresponding sides is equal

$\frac{AB}{CD}=\frac{AC}{CE}$

substitute and solve for CE

$\frac{6}{4}=\frac{10}{CE}\\ \\CE=4*10/6\\ \\CE=\frac{20}{3}\ units$

step 3

Find the length of segment AE

AE=AC+CE

substitute the values

$AE=10\ units+\frac{20}{3}\ units=\frac{50}{3}\ units$

6 0

3 years ago

Recall that Rn denotes the right-endpoint approximation using n rectangles, Ln denotes the left-endpoint approximation using n r

pogonyaev

Answer:

$R_5=1.12$

Step-by-step explanation:

We want to calculate the right-endpoint approximation (the right Riemann sum) for the function:

$f(x)=x^2+x$

On the interval [-1, 1] using five equal rectangles.

Find the width of each rectangle:

$\displaystyle \Delta x=\frac{1-(-1)}{5}=\frac{2}{5}$

List the x-coordinates starting with -1 and ending with 1 with increments of 2/5:

-1, -3/5, -1/5, 1/5, 3/5, 1.

Since we are find the right-hand approximation, we use the five coordinates on the right.

Evaluate the function for each value. This is shown in the table below.

Each area of each rectangle is its area (the y-value) times its width, which is a constant 2/5. Hence, the approximation for the area under the curve of the function f(x) over the interval [-1, 1] using five equal rectangles is:

$\displaystyle R_5=\frac{2}{5}\left(-0.24+-0.16+0.24+0.96+2)= 1.12$

3 0

3 years ago

Can you use elimination to solve a system of linear equations with three equations

crimeas [40]

You can use elimination to solve systems of equations with 3 equations. I know how to solve systems of equatons with 3 equations, but I use a different process, I don't know how to use the elimination method.

7 0

3 years ago

Read 2 more answers

A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa

PSYCHO15rus [73]

Answer:

A) $\dfrac{11}{23}$

B) $\dfrac{88}{345}$

Step-by-step explanation:

A survey of 46 college athletes found that

24 played volleyball,
22 played basketball.

A) If we pick one athlete survey participant at random, the probability they play basketball is

$P_1=\dfrac{22}{46}=\dfrac{11}{23}$

B) If we pick 2 athletes at random (without replacement),

the probability we get one volleyball player is $\dfrac{24}{46}=\dfrac{12}{23};$
the probability we get another basketball player is $\dfrac{22}{45}$ (only 45 athletes left).

Thus, the probability we get one volleyball player and one basketball player is

$P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}$

5 0

4 years ago