Ask question

Ask question

All categories

2 years ago

6

Could someone help out ? thanks

1 answer:

RUDIKE [14]2 years ago

6 0

Since the area of the walkway is not given, it's substitution in the answer is x. You do have the measurements of the pool though. Area=length times width. 36 times 18 is 648. The area of the concrete is x-648 ft, because the pool is not apart of the concrete. The area of the walkway can be expressed by X-648 ft.

You might be interested in

Find z such that 7.5% of the standard normal curve lies to the left of z.

sergeinik [125]

A probability of 0.075 corresponds to a z-score of about $z\approx-1.44$ , i.e.

$\mathbb P(Z$

3 0

3 years ago

Which statement BEST compares the growth of the two functions as x approaches oo?

Karo-lina-s [1.5K]

Answer:

b girl main daal atom answer

Step-by-step explanation:

kas goal feel keep PCL pal PCL

6 0

2 years ago

If a car is traveling at 65 miles per hour, then how far does it travel in 2 hours?

frez [133]

65 miles per hour means that for each hour, it goes 65 miles

65 miles*2 hours= 130 miles

Final answer: 130 miles

5 0

3 years ago

Read 2 more answers

Please someone help me. Thank you.

KiRa [710]

Answer:

Step-by-step explanation:

take x degree as reference angle

using tan rule

tan x=opposite/adjacent

tan x=6.1/5.1

tan x=1.19

x= $tan^{-1}$ 1.19

x=49.95

x=50 degree

3 0

3 years ago

Tienes un triángulo equilátero con lados de 10cm, lo divides en dos partes de manera horizontal tal que el área de ambos figuras

Tanya [424]

Answer:

x = 10/√2 ≈ 7.07

Step-by-step explanation:

Comenzaremos por dividir el triángulo en dos partes y definir H, como en la figura adjunta.

Aplicando el teorema de Tales, sabemos que:

$\dfrac{l/2}{H}=\dfrac{x/2}{h}\\\\\\\dfrac{l}{H}=\dfrac{x}{h}$

También sabemos que, dado que el tirángulo menor es la mitad que el triángulo mayor, la relación entre áreas es:

$\dfrac{A}{A_x}=\dfrac{lH/2}{xh/2}=\dfrac{lH}{xh}=2$

Dado que formamos dos triángulos rectángulos, podemos despejar el valor de H como:

$(l/2)^2+H^2=l^2\\\\H^2=l^2-(l/2)^2=10^2-5^2=100-25=75\\\\H^2=\sqrt{75}$

Podemos entonces despejar x de la siguiente manera:

$h=\dfrac{H}{l}\cdot x=\dfrac{lH}{2x}\\\\\\2x^2\dfrac{H}{l}=lH\\\\\\2x^2=l^2\\\\\\x=\dfrac{l}{\sqrt{2}}=\dfrac{10}{\sqrt{2}}\approx7.07$

6 0

3 years ago