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jek_recluse [69]
2 years ago
11

Yolanda found that her hair

Mathematics
1 answer:
Lisa [10]2 years ago
6 0

Answer:

7.05 inches

Step-by-step explanation:

h=7.5+.59(5)-3.4

h=7.5+2.95-3.4

h=10.45-3.4

h=7.05

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What is 246,588 rounded to the ten thousands place?
Novosadov [1.4K]

246,588 rounded to the nearest 10 thousands place is 250,000

8 0
2 years ago
Given: x - 5 &gt; -2. <br><br> Choose the solution set.
slega [8]
<span>x - 5 > -2
x > -2 + 5
x > 3

x </span>∈ (3; +∞)
3 0
3 years ago
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What is the slope of the line that passes through the points (-2,-5) and
Gemiola [76]

We can use the points (-2, -5) and (18, -5) to solve.

Slope formula: y2-y1/x2-x1

= -5-(-5)/18-(-2)

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Best of Luck!

8 0
3 years ago
One more question on this test guys. please answer as fast as possible. If you do,
GaryK [48]

1. The given rectangular equation is x=2.

We substitute x=r\cos \theta.

r\cos \theta=2

Divide through by \cos \theta

r=\frac{2}{\cos \theta}

r=2}\sec \theta

\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

We use the relation r^2=x^2+y^2 and y=r\sin \theta

This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

4. We have x=\sqrt{3}y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

\theta=\frac{\pi}{6}

\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}

5. We have x=y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta

This implies that;

\tan \theta=1

\theta=\frac{\pi}{4}

\boxed{x=y\to \theta=\frac{\pi}{4}}

6 0
3 years ago
A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o
vivado [14]

Since \rho=\dfrac mV (density = mass/volume), we can get the mass/weight of the liquid by integrating the density \rho(x,y,z) over the interior of the tank. This is done with the integral

\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx

which is more readily computed in cylindrical coordinates as

\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}

7 0
3 years ago
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