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Masja [62]
3 years ago
5

What is the circumference of a circle with an area of 50.24

Mathematics
1 answer:
Pepsi [2]3 years ago
3 0

Answer: C=25.12

Step-by-step explanation:

\displaystyle\ \Large \boldsymbol{} if  \ \ equal   \  \pi =3,14  \ \  then \\\\\pi r^2=3,14r^2\\\\3,14r^2=50,24 \\\\ r^2=16 \\\\ r=4  \Longrightarrow C=  2\pi r=2\cdot 3,14\cdot 4=25.12 \\\\\\ Answer : C=25.12

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If RSTU is a rhombus, find the measure of angle RUT.
Schach [20]

<u>Finding x:</u>

We know that the diagonals of a rhombus bisect its angles

So, since US is a diagonal of the given rhombus:

∠RUS = ∠TUS

10x - 23 = 3x + 19                      [replacing the given values of the angles]

7x - 23 = 19                                [subtracting 3x from both sides]

7x = 42                                       [adding 23 on both sides]

x = 6                                           [dividing both sides by 7]

<u>Finding ∠RUT:</u>

We can see that:

∠RUT = ∠RUS + ∠TUS

<em>Since we are given the values of ∠RUS and ∠TUS:</em>

∠RUT = (10x - 23) + (3x + 19)

∠RUT = 13x - 4

<em>We know that x = 6:</em>

∠RUT = 13(6)-  4

∠RUT = 74°

8 0
3 years ago
What is the slope of the line through points (0,6) and (5,-4)
anastassius [24]
The slope between the points (x1,y1) and (x2,y2) is
slope=(y2-y1)/(x2-x1)

(0,6) and (5,-4)
x1=0
y1=6
x2=5
y2=-4

slope=(-4-6)/(5-0)=-10/5=-2
slope=-2
7 0
3 years ago
Colin is skateboarding and is riding a rail. The rail is attached to two poles. The first pole is 80 cm tall and the
Vinvika [58]

Answer:

\sqrt{4.49m}

Step-by-step explanation:

the unit of length is meter,

so, 80cm=0.8cm, 10cm=0.1m

the length of the rail=\sqrt{(0.8-0.1)^2+2^2}

                                  =\sqrt{4.49m}

5 0
2 years ago
Find the slope in fraction form
nlexa [21]

Answer:

-1/3

Step-by-step explanation:

Taking two points on the graph (I took (0,9) and (6,7)

Slope=(y2-y1)/(x2-x1)

(7-9)/(6-0)=-2/6

=-1/3

7 0
4 years ago
Read 2 more answers
Find the volume of the solid formed by rotating the region bounded by the given curves about the indicated axis. Y = 1x, y = 1,
Goryan [66]

The disk method will only involve a single integral. I've attached a sketch of the bounded region (in red) and one such disk made by revolving it around the y-axis.

Such a disk has radius x = 1/y and height/thickness ∆y, so that the volume of one such disk is

π (radius) (height) = π (1/y)² ∆y = π/y² ∆y

and the volume of a stack of n such disks is

\displaystyle V_n = \sum_{i=1}^n \pi {y_i}^2 \Delta y

where y_i is a point sampled from the interval [1, 5].

As we refine the solid by adding increasingly more, increasingly thinner disks, so that ∆y converges to 0, the sum converges to a definite integral that gives the exact volume V,

\displaystyle V = \lim_{n\to\infty} V_n = \int_1^5 \frac{\pi}{y^2} \, dy

V = -\dfrac\pi y\bigg|_{y=1}^{y=5} = \boxed{\dfrac{4\pi}5}

8 0
2 years ago
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