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3 years ago

Express 180 as a product of its prime factors. 3. 180-2

Mathematics

1 answer:

mariarad [96]3 years ago

6 0

Answer:

180=2²×3²×5

Step-by-step explanation:

180=18×10=(9×2)×(2×5)=(3×3×2)×(2×5)=

2²×3²×5

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In DEF, DE = 15, and m angle F=32 Find EF to the nearest tenth.

iren [92.7K]

the answer would be 24.0.

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3 years ago

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

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

The equation of the graphed line in slope point form using the point (2,-1 ) is blank and it’s equation in slope intercept is bl

valkas [14]

Answer:

$y=\frac{1}{4}x-\frac{3}{2}$ (Slope intercept form)

$y+1=\frac{1}{4}(x_-2)$ (Slope point form)

Step-by-step explanation:

The slope of the line can be calculated as following:

$m=\frac{-1-(-2)}{2-(-2)}=1/4$

The the equation of the line in slope point form is:

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

Where ( $x_1$ , $y_1$ ) is a point of the line.

Then you must susbtitute the point given and the slope and you obtain:

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

The intercept form is:

$y=mx+b$

Where b is the y-intercept.

Solve for y from the equation of the line in slope point form obtained above, then:

$y+1=\frac{x}{4}-\frac{2}{4}\\y=\frac{1}{4}x-\frac{2}{4}-1\\y=\frac{1}{4}x-\frac{3}{2}$

8 0

3 years ago

Please will anyone help me out with this math problem? I will give them a crown.

Aneli [31]

Answer:

a. 450 b. 10078.9097

Step-by-step explanation:

a. go with the formula W=FD

F= force

W= work done

D= distance

b. go with the formula F=mgcos(x)

M= mass

G= gravity constant (9.8)

then get to the normal formula of W=FD after getting the Force

4 0

3 years ago

999+6666+8888+66654+66=

scoundrel [369]

Answer:

83273

Step-by-step explanation:

8 0

3 years ago