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3 years ago

What is the value of x? A.35 B.60 C.155 D.25

Mathematics

2 answers:

vivado [14]3 years ago

6 0

For this question, since there is 180 in a triangle, subtract 60+85 to get A.35

Elena L [17]3 years ago

6 0

Answer: the answer is 35 degrees

Step-by-step explanation:

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can someone help me with this I have to do like 18 other questions so if you can help me with those I would appreciate it!

GenaCL600 [577]

Answer:

see below

Step-by-step explanation:

1.)

(0, 4)

(1, 7)

$\frac{7-4}{1-0} = 3$

y = 3x + b

(4) = 3(0) + b

b = 4

y = 3x + 4

Rate of change: 3 initial value: 4

2.)

(0, -5)

(1, -3)

$\frac{-3-(-5)}{1-0} = 2$

y = 2x + b

(-5) = 2(0) + b

b = -5

Rate of change: 2 initial value: -5

4 0

2 years ago

Graph the function f(x)=−14x−2. Use the line tool and select two points to graph.

Akimi4 [234]

Answer:

You should do this on your own but if you consider the way I graph it, the slope should be going downwards and in the middle of the graph, the y-intercept should be two units down. Each unit for the x line should be 14 units apart as stated in the function.

Step-by-step explanation:

Hopefully this helps ^^

7 0

3 years ago

Read 2 more answers

Find m∠MON pls______

Wittaler [7]

Answer:

57°

Step-by-step explanation:

∠LOM and ∠MON are complementary angles, so they add up to 90.

This means that:

3x-15 + 5x-23 = 90 =>

8x - 38 = 90 =>

x = 128/8 = 16

So m∠MON = 5*16 - 23 = 57°

5 0

3 years ago

5p−3p+9+p+3? pls im failingg

Triss [41]

Answer:

3p + 12

Step-by-step explanation:

5p -3p + 9 + p + 3

Move around the parts of the equation

5p -3p + p + 9 + 3

Add positive 9 to positive 3

5p -3p + p + 12

Add positive p to negative 3p

5p -2p + 12

Combine positive 5p and negative 2p together

3p + 12

I Hope That This Helps! :)

7 0

3 years ago

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

3 0

3 years ago