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3 years ago

Help please NO LINKS Find the angle

Mathematics

1 answer:

algol [13]3 years ago

6 0

Answer:

x = 50.8°

Step-by-step explanation:

We are given:

1.) Angle x

2.) Side opposite to angle x

3.) Hypotenuse

Therefore to find the angle x, we will use the sin inverse:

<em>Sin(x) = Opposite / Hypotenuse</em>

Sin(x) = 6.2/8

x = Sin⁻¹(6.2/8)

Hope this helps!

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Solve the system of equations using subtraction.<br> 4x - 2y = 10<br> 4x + y = 19

Alex777 [14]

Answer:

y=3 x=4

Step-by-step explanation:

4x - 2y = 10

-3y=-9

the 4x canceled out.

y=3

plug in y into one of the equations

4x + 3 = 19

4x=16

x=4

Hope this helps!

If not, I am sorry.

3 0

2 years ago

Read 2 more answers

The product of -6 and a number is at least -72.

sesenic [268]

-6x>= -72. {x| 12>=x}

-72 (less than or equal to sign) -6x (product means multiply right?) all solutions is the value 12 and any value less

6 0

3 years ago

Find the center of mass of the wire that lies along the curve r and has density =4(1 sin4tcos4t)

dolphi86 [110]

The mass of the wire is found to be 40π√2 units.

To calculate the mass of the wire which runs along the curve r ( t ) with the density function δ=5.

The general formula is,

Mass = $\int_a^b \delta\left|r^{\prime}(t)\right| d t$

To find, we must differentiate this same given curve r ( t ) with respect to t to estimate |r'(t)|.

The given integration limits in this case are a = 0, b = 2π.

Now, as per the question;

The equation of the curve is given as;

r(t) = (4cost)i + (4sint)j + 4tk

Now, differentiate this same given curve r ( t ) with respect to t.

$\begin{aligned}\left|r^{\prime}(t)\right| &=\sqrt{(-4 \sin t)^2+(4 \cos t)^2+4^2} \\&=\sqrt{16 \sin ^2 t+16 \cos ^2 t+16} \\&=\sqrt{16\left(\sin t^2+\cos ^2 t\right)+16}\end{aligned}$

Further simplifying;

$\begin{aligned}&=\sqrt{16(1)+16} \\&=\sqrt{16+16} \\&=\sqrt{32} \\\left|r^{\prime}(t)\right| &=4 \sqrt{2}\end{aligned}$

Now, use integration to find the mass of the wire;

$\begin{aligned}&=\int_a^b \delta\left|r^{\prime}(t)\right| d t \\&=\int_0^{2 \pi} 54 \sqrt{2} d t \\&=20 \sqrt{2} \int_0^{2 \pi} d t \\&=20 \sqrt{2}[t]_0^{2 \pi} \\&=20 \sqrt{2}[2 \pi-0] \\&=40 \pi \sqrt{2}\end{aligned}$