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3 years ago

Help help please please

Mathematics

1 answer:

Ahat [919]3 years ago

7 0

Answer:

1. A

2. C

Step-by-step explanation:

I could be wrong, that's my best guess. Im so sorry if im wrong, have a good day

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2 years ago

The volume of 100 drops of a liquid is 0.1 fluid ounce. What is the volume of 1000 drops?

Ket [755]

Hey there!

100 drops of the liquid is 0.1 fluid ounces.

1,000 is equal to 10 x 100.

To find the volume of 1,000 drops of liquid, we must multiply the volume of 100 drops, which is 0.1, by 10.

0.1 x 10 = 1

Therefore, the volume of 1,000 drops of this liquid is 1 fluid ounce.

Hope this helps!

6 0

3 years ago

What is the equation of the quadratic graph with a focus of (6, 0) and a directrix of y = −10 ?

Mademuasel [1]

Answer:

$(x-6)^{2}=20(y+5)^{2}$

Step-by-step explanation:

The standard for for the equation of a parabola is $(x-h)^{2}=4p(y-k)$

The focus is given as $(h, k+p)$

The directrix is given by $y=k-p$

Comparing focus given as $(6,0)$ to formula of focus $(h,k+p)$ , we see that $h=6$ and $k+p=0$
Comparing directrix given as $y=-10$ to formula of directrix $y=k-p$ , we can write $k-p=-10$

We can use the two system of equations [ $k+p=0$ and $k-p=-10$ ] to solve for $k$ and $p$ .

Adding the two equations gives us:

$2k=-10\\k=\frac{-10}{2}\\k=-5$

Using this value of $k$ , we can find $p$ by plugging this value in either equation. Let's put it in Equation 1. We have:

$k-p=-10\\-5-p=-10\\-5+10=p\\p=5$

Now, that we know $h=6$ , $k=-5$ , and $p=5$ , we can plug these values in the standard form equation to figure out the parabola's equation. Doing so and rearranging gives us:

$(x-6)^{2}=4(5)(y-(-5))^{2}\\(x-6)^{2}=20(y+5)^{2}\\$

This is the equation of the parabola with focus $(6,0)$ and directrix $y=-10$

3 0

3 years ago

Read 2 more answers

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3>Question 1</h3>

Method 1 - modelling as a quadratic with restricted domain

Assuming that the points given on the graph are points that the curve passes through, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

Equation 1 (-4, 2)

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

Equation 2 (-1, 4)

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

Equation 3 (4, 6)

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

Method 2 - modelling as a square root function

Assuming that the points given on the graph are points that the curve passes through, and the x-intercept should be included, we can model this curve as a square root function.

Given points: