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2 years ago

All of these value from smallest to largest

Mathematics

1 answer:

ollegr [7]2 years ago

3 0

You didn't put the values..??

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does anyone have the unit 3 expressions, equations, and inequalities unit test: B answer key. i really really need it. this is a

Vilka [71]

Answer:

sorry but you can write question instead of writing that because some of you may know that but others donot knew it

5 0

2 years ago

Simplify 5^3/5^7 leaving your anwser in index form

Wewaii [24]

Steps:

Simplify:

5³/5⁷

= 5³/5⁷

= 5*5*5/ 5*5*5*5*5*5*5

=1/5⁴

=1/625 (Decimal: 0.0016)

Steps by Step:

5³/5⁷

=125/5⁷

=125/78125

=1/625

Answer: =1/625 (Decimal: 0.0016)

Please mark brainliest

<em><u>Hope this helps.</u></em>

6 0

2 years ago

LAST QUESTION please just help me out please

Maru [420]

Shiii still here for point
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6 0

2 years ago

If the product of two consecutive odd integers is decreased by one-third the lesser integer, the result is 150. find the integer

dlinn [17]

X(x+2)-(1/3)x=150
x²+2x-(1/3)x-150=0
x²+(5/3)x-150=0
3x²+5x-450=0
this does not produce an integer, because b²-2ac=5²+4*3*450=5425, and √5425=73.6546 is not an integer.
Please double check your question.

6 0

2 years ago

A 12 ft. ladder is leaning against a house. The ladder makes a 60° angle with the ground. Use special right triangles to find ho

OlgaM077 [116]

Given:

The length of the ladder = 12 ft

The angle of ladder with ground = 60 degrees

To find:

How far up the building the ladder will reach.

Solution:

Using the given information draw a figure as shown below.

We need to find the vertical distance between the top of ladder and the ground.

Let x be the required distance.

In a right angle triangle,

$\sin \theta=\dfrac{Perpendicular}{Hypotenuse}$

In the below triangle ABC,

$\sin A=\dfrac{BC}{AC}$

$\sin 60^\circ=\dfrac{x}{12}$

$\dfrac{\sqrt{3}}{2}=\dfrac{x}{12}$

Multiply both sides by 12.

$\dfrac{\sqrt{3}}{2}\times 12=x$

$6\sqrt{3}=x$

Therefore, the ladder will reach $6\sqrt{3}$ ft far up the building.

4 0

2 years ago