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3 years ago

Helppppp please !!!!!

Mathematics

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

The answer is A) 1/8^2

Step-by-step explanation:

Both of the equations simplified equal 1/64.

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PLSS give me the answer for the B and C!! ( there two different questions)

Rzqust [24]

Answer:

Circumference Formula :

$2\pi r$

$\frac{120}{360}$

This is the ratio we can work out:

$\frac{given \: angle}{revolution} = ratio$

$arc = ratio \times circumference$

$22 = \frac{120}{360} \times 2 \times \frac{22}{7} \times r$

$22 = \frac{44}{21} x$

$x = 22 \div \frac{41}{21}$

$x = \frac{21}{2}$

5 0

3 years ago

(05.05 MC) The length of a rectangl e is shown below: A B 4 31 2 1 -6 -5 4 3 2 0 2 -1 -2 -3 4 -5 If the area of the rectangle to

kolbaska11 [484]

Answer:

ポイントが欲しいだけなのかわからない

Step-by-step explanation:

ポイントが欲しいだけなのかわからない

4 0

3 years ago

A 30° 60° 90° triangle is shown below. Find the length of the side labeled y.

bulgar [2K]

Answer:

Part 1) $y=3\ units$

Part 2) The length of the hypotenuse is 6 units

Step-by-step explanation:

Part 1) we know that

In the right triangle of the figure

$sin(60^o)=\frac{y}{2\sqrt{3}}$ ----> by SOH (opposite side divided by the hypotenuse)

solve for y

$y=sin(60^o)2\sqrt{3}$

Remember that

$sin(60^o)=\frac{\sqrt{3}}{2}$

substitute

$y=(\frac{\sqrt{3}}{2})2\sqrt{3}$

$y=3\ units$

Part 2)

Let

h ----> the length of the hypotenuse

we know that

In the right triangle of the figure

$sin(45^o)=\frac{3\sqrt{2}}{h}$ ----> by SOH (opposite side divided by the hypotenuse)

$h=\frac{3\sqrt{2}}{sin(45^o)}$

Remember that

$sin(45^o)=\frac{\sqrt{2}}{2}$

substitute

$h=3\sqrt{2}:\frac{\sqrt{2}}{2}$

$h=6\ units$

7 0

3 years ago

1.N% of 40 is 10. Solve for N. Just put the number.

Serhud [2]

Step-by-step explanation:

N% of 40=10

N/100×40=10

N=10×100/40=25

60%of homework=45min

1%of homework=45/60=3/4min

100%of homework =3/4 ×100=75 min

7 0

3 years ago

yz = 7ln(x + z), (0, 0, 1) (a) the tangent plane Correct: Your answer is correct. (b) parametric equations of the normal line to

Alika [10]

Answer:

The equation of tangent plane is 7x - y + 7z - 7 = 0

Parametric equation of normal line

x = 7 t , y=-t , z=1+7 t

Step-by-step explanation:

Equation of tangent

fₓ (x₀ , y₀ , z₀) (x-x₀) + fy (x₀ , y₀ , z₀) (y-y₀) +fz(x₀ , y₀ , z₀)(z-z₀)=0 (1)

From taking derivation we get

fₓ (x₀ , y₀ , z₀) = 7

fy (x₀ , y₀ , z₀)= -1

fz(x₀ , y₀ , z₀) = 7

putting these value in equation 1

(7) (x-0) + (-1)(y-0) + 7(z-1)=0

7x - y + 7z - 7 = 0

The equation of tangent plane is 7x - y + 7z - 7 = 0

b) Parametric equation

x=x +f ₓ (P)t , y = y₀ +f y (P) t , z=z₀ +f z (P) t

x=0 +7 t , y =0+(-1) t , z=1+7 t

x=7 t ,y=-t , z= 1+7 t

Parametric equation of normal line

x = 7 t , y=-t , z=1+7 t

7 0

4 years ago

Helppppp please !!!!!​

Helppppp please !!!!!