Evaluate the function y = 17 – x when x =

Simplify 81 power 5/4 A. 3 B. 243 c. 6,561 d. 59,049

VARVARA [1.3K]

Answer:

B

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A square park measures 170 feet along each side. Two paved paths run from each corner to the opposite corner and extend 3 feet i

Cerrena [4.2K]

Answer:

The total area, in square feet, taken by the paths is 2,004

Step-by-step explanation:

see the attached figure with lines to better understand the problem

I can divide the figure into four right triangles, one small square and four rectangles

step 1

Find the area of the right triangle of each corner of the path

The area of the triangle is

$A=(1/2)(b)(h)$

substitute the given values

$A=(1/2)(3)(3)=4.5\ ft^2$

step 2

Find the hypotenuse of the right triangle

Applying Pythagoras Theorem

Let

d -----> hypotenuse of the right triangle

$d^{2}=3^{2}+3^{2}$

$d^{2}=18$

$d=\sqrt{18}\ ft$

simplify

$d=3\sqrt{2}\ ft$

The hypotenuse of the right triangle is equal to the width of the path

step 2

Find the area of the small square of the path

The area is

$A=b^{2}$

we have

$b=3\sqrt{2}\ ft$ ----> the width of the path

substitute

$A=(3\sqrt{2})^{2}$

$A=18\ ft^2$

step 3

Find the length of the diagonal of the square park

Applying Pythagoras Theorem

Let

D -----> diagonal of the square park

$D^{2}=170^{2}+170^{2}$

$D^{2}=57,800$

$D=\sqrt{57,800}\ ft$

simplify

$D=170\sqrt{2}\ ft$

step 4

Find the height of each right triangle on each corner

The height will be equal to the width of the path divided by two, because is a 45-90-45 right triangle

$h=1.5\sqrt{2}\ ft$

step 5

Find the area of each rectangle of the path

The area of rectangle is $A=LW$

we have

$W=3\sqrt{2}\ ft$ ----> width of the path

Find the length of each rectangle of the path

$L=(D-2h-d)/2$

where

D is the diagonal of the park

h is the height of the right triangle in the corner

d is the width of the path (length side of the small square of the path)

substitute the values

$L=(170\sqrt{2}-2(1.5\sqrt{2})-3\sqrt{2})/2$

$L=(170\sqrt{2}-3\sqrt{2}-3\sqrt{2})/2$

$L=(164\sqrt{2})/2$

$L=82\sqrt{2}\ ft$

Find the area of each rectangle of the path

$A=LW$

we have

$W=3\sqrt{2}\ ft$

$L=82\sqrt{2}\ ft$

substitute

$A=(82\sqrt{2})(3\sqrt{2})$

$A=492\ ft^2$

step 6

Find the area of the paths

Remember

The total area of the paths is equal to the area of four right triangles, one small square and four rectangles

so

substitute

$A=4(4.5)+18+4(492)=2,004\ ft^2$

therefore

The total area, in square feet, taken by the paths is 2,004

3 0

3 years ago

This my last one help

Lerok [7]

Answer:

A2 = 120

A3 = 60

A5 = 60

A6 = 120

Step-by-step explanation:

Angle 1 and 2 are supplementary, so A2 = 180 - 60, which is 120.

Angle 1 and 3 are vertical angles, which are always the same measure.

Angle 5 is alternate interior angles with 3, which means it is also 60.

Angle 6 is alternate interior angles with 4, which means it is 120.

3 0

3 years ago

5TH GRADE MATH HELP What is 4.873 rounded to the nearest hundredth? A. 4.87 B. 4.88 C. 4.9 D. 5

timama [110]

the answer is a becuse you tacke that little dot and you slid it to the left tow times

4 0

4 years ago

Read 2 more answers

Suppose we have a right triangle with legs of length a and b and hypotenuse of length c. Suppose b=3 and c=5. Then a= , For the

ANTONII [103]

Answer:

Length of right-angle triangle 'a' = 4

b)

$sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$

$cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$

$tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$

Step-by-step explanation:

Step(i):-

Given b = 3 and hypotenuse c = 5

Given ΔABC is a right angle triangle

By using pythagoras theorem

c² = a² + b²

⇒ a² = c² - b²

⇒ a² = 5²-3²

=25 - 9

a² = 16

⇒ a = √16 = 4

The sides of right angle triangle a = 4 ,b = 3 and c = 5

Step(ii):-

$sin(A) = \frac{opposite side}{Hypotenuse} = \frac{a}{c} = \frac{4}{5}$

$cos(A) = \frac{Adjacent side}{Hypotenuse} = \frac{b}{c} = \frac{3}{5}$

$tan(A) = \frac{opposite side}{Adjacent side} = \frac{a}{b} = \frac{4}{3}$

7 0

3 years ago

Evaluate the function y = 17 – x when x = –6