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2 years ago

Priya has picked 1 1/2 cups of raspberries which is enough for 3/4 of the cake how many cups does she need for the whole cake

Mathematics

1 answer:

Zina [86]2 years ago

3 0

Answer:

So Priya need $2$ cups of raspberries for the whole cake.

Step-by-step explanation:

Given data;

$1\frac{1}{2}=\frac{3}{2}$ cups of raspberries which is enough for $\frac{3}{4}$ of the cake.

For making $\frac{3}{4}$ of the cake need $\frac{3}{2}$ cups of raspberries.

For making $1$ cake need $\frac{\frac{3}{2}}{\frac{3}{4}}$ cups of raspberries.

For making $1$ cake need $\frac{3}{2} \times \frac{4}{3}=2$ cups of raspberries.

∴ Priya need $2$ cups of raspberries for the whole cake.

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HELP PLEASE! Can you please explain so I can understand how it was completed?

nlexa [21]

Answer:

(3 square root of 2 , 135°), (-3 square root of 2 , 315°)

Step-by-step explanation:

Hello!

We need to determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°.

We know that the polar coordinate system is a two-dimensional coordinate. The two dimensions are:

The radial coordinate which is often denoted by r.

The angular coordinate by θ.

So we need to find r and θ. So we know that:

$r=\sqrt{x^{2}+y^{2}}$ (1)

x = rcos(θ) (2)

x = rsin(θ) (3)

From the statement we know that (x, y) = (3, -3).

Using the equation (1) we find that:

$r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+(-3)^{2}} = 3\sqrt{2}$

Using the equations (2) and (3) we find that:

3 = rcos(θ)

-3 = rsin(θ)

Solving the system of equations:

θ= -45

Then:

r = 3\sqrt{2}[/tex]

θ= -45 or 315

Notice that there are two feasible angles, they both have a tangent of -1. The X will take the positive value, and Y the negative one.

So, the solution is:

(3 square root of 2 , 135°), (-3 square root of 2 , 315°)

4 0

2 years ago

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8

Gekata [30.6K]

First, let $(x,y)$ be a point in our parabola. Since we know that the focus of our parabola is the point (0,8), we are going to use the distance formula to find the distance between the two points:
$\sqrt{(x-0)^{2}+(y-8)^{2}}$

Next, we are going to find the distance between the directrix and the point in our parabola. Remember that the distance between a point (x,y) of a parabola and its directrix, $y=c$ , is: $|y-c|$ . Since our directrix is y=-8, the distance to our point will be:
$|y-(-8)|$
$|y+8|$

Now, we are going to equate those two distances, and square them to get rid of the square root and the absolute value:
$\sqrt{(x-0)^{2}+(y-8)^{2}}=|y+8|$
$(\sqrt{(x-0)^{2}+(y-8)^{2}})^{2}=|y+8|^{2}$
$(x-0)^{2}+(y-8)^{2}=y+8$

Finally, we can expand and solve for $y$ :
$x^2+y^2-16y+64=y^2+16y+64$
$x^{2}-16y=16y$
$32y=x^{2}$
$y= \frac{1}{32} x^2$

We can conclude that t<span>he standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8 is </span> $y= \frac{1}{32} x^{2}$