REPOST HELP PLEASEEE I KNOW THIS ISNT MATH BUT NO ONE IS SEEING IT

Consider two circular swimming pools. Pool A has a radius of 22 feet, and Pool B has a diameter of 13.6 meters. Complete the des

Black_prince [1.1K]

Answer:

See below.

Step-by-step explanation:

Diameter of pool A = 2 * radius = 2 * 22

= 44 feet

= 44 * 0.305 = 13.42 meters.

The diameter of Pool B ( 13. 6 m) is greater than the diameter of Pool A.

The circumference of Pool A is 13.42 * pi = 42.16 m and that of B is 42.73 m, so that the circumference of Pool B is greater than that of Pool A by

0.57 m.

4 0

3 years ago

Find the values of x and y that make these triangles congruent by the HL theorem

GarryVolchara [31]

Answer:

Step-by-step explanation:

As due to congruency,

x + 3 = 3y

[By putting the values of x = 3 and y = 2]

=> 3 + 3 = 3 × 2

=> 6 = 6

and,

x = y + 1

[By putting the values of x = 3 and y = 2]

=> 3 = 2 + 1

=> 3 = 3

Hence, proved

6 0

3 years ago

A. 60°

Tju [1.3M]

Answer:

60° is the value of y....

7 0

3 years ago

Read 2 more answers

A rectangular shoebox has a volume of 728 cubic inches. The base of the shoebox measures 8 inches by 6.5 inches. How long is the

siniylev [52]

Okay to solve this you need to divide the volume to the two measurements
728 ÷ 8 = 91 ÷ 6.5 = 14

7 0

3 years ago

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati

murzikaleks [220]

Answer:

$x-4y+4=0$

$f(x)=\sqrt x$ and x=4

Step-by-step explanation:

We are given that a curve

$y=\sqrt x$

We have to find the equation of tangent at point (4,2) on the given curve.

Let y=f(x)

Differentiate w.r.t x

$f'(x)=\frac{dy}{dx}=\frac{1}{2\sqrt x}$

By using the formula $\frac{d(\sqrt x)}{dx}=\frac{1}{2\sqrt x}$

Substitute x=4

Slope of tangent

$m=f'(x)=\frac{1}{2\sqrt 4}=\frac{1}{2\times 2}=\frac{1}{4}$

In given question

$m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$

$\frac{1}{4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}$

By comparing we get a=4

Point-slope form

$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (4,2)

$y-2=\frac{1}{4}(x-4)$

$4y-8=x-4$

$x-4y-4+8=0$

$x-4y+4=0$

8 0

3 years ago