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4 years ago

I need help it has to be a mixed number, three numbers Please and Thank You.

Mathematics

1 answer:

Umnica [9.8K]4 years ago

5 0

You basically do whatever is bigger than the one your comparing it to that’s how I do it in my school

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Find the elapsed time between the pair of times. <br><br> from 4:49 p.m to 5:18 a.m

Roman55 [17]

I got 12 hours and 31 minutes.

5 0

4 years ago

The value 0.01 gram is equivalent to 1 hectogram True or False

klemol [59]

Answer:

This is false statement.

Step-by-step explanation:

⇒ 1 Gram = 0.01 hectogram

⇒ 1 hectogram = 100 gram

⇒ 0.01 gram = 0.0001 hectogram

Thus the given statement is false.

3 0

3 years ago

There are three choices of jellybeans: lemon, strawberry, and orange. If the probability of getting a lemon is 5/10 and the prob

geniusboy [140]

Answer:

3/10 chance of getting orange

Step-by-step explanation:

1/5 = 2/10

10/10 - 2/10 - 5/10 = 3/10 chance of getting orange

7 0

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

Solve 10y4+6y2-12y3-5y4

Olin [163]

Answer:

(5y² - 12y + 6) y²

Step-by-step explanation:

i'm going to assume that you mean to simplify the expression:

10y⁴+6y²-12y³-5y⁴ (group like terms)

= (10y⁴- 5y⁴) - 12y³ + 6y²

= 5y⁴ - 12y³ + 6y² (factor out y²)

= (5y² - 12y + 6) y²

4 0

4 years ago

I need help it has to be a mixed number, three numbers Please and Thank You.​

I need help it has to be a mixed number, three numbers Please and Thank You.