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3 years ago

Write 10.895.73 in word form and 11.273. 26 in word form

Mathematics

2 answers:

bija089 [108]3 years ago

6 0

I don't think a decimal number could have 2 decimal points.

denis-greek [22]3 years ago

6 0

You have some typos, first off. are the periods decimal points? Or commas, second. the final two numbers seperated by the periods, cannot be two numbers if there are numbers in front of it. It has to be three, if you fix those typos then the answer would be:
1. one million, eighty-nine thousand, five hundred seventy three.
2. one million, one hundred twenty-seven thousand, three hundred twenty-six
Hope this helps :)

You might be interested in

The longest leg of a right triangle is 12 inches. The hypotenuse is 3 inches longer then twice the shortest leg. How long is the

Llana [10]

Answer:

Shortest Leg: 5 inches, Hypotenuse: 13 inches

Step-by-step explanation:

A right triangle follows the pythagorean theorem, a^2+b^2=c^2. C is the hypotenus. A combination found using this is 5, 12, and 13. To double check this, we know that the hypotenus is 2x+3, where x is the smallest leg.

6 0

3 years ago

A seagull was flying about sea level looking to catch a fish below sea level. The elevation above sea level was 125 1/2 feet. He

frez [133]

Answer:

The seagull's position halfway is 85.5 feet.

Step-by-step explanation:

The seagull is at 125 1/2 feet above the sea level.

The fish is at 45 1/2 feet below the sea level.

The distance between the seagull and fish = seagull's distance above the sea level + fish's distance below the sea level

= 125 1/2 + 45 1/2

= $\frac{251}{2}$ + $\frac{91}{2}$

= $\frac{342}{2}$

= 171

The distance between the seagull and fish is 171 feet.

Since the seagull stops halfway, then;

seagull's position = $\frac{171}{2}$

= 85.5

The seagull's position halfway is 85.5 feet.

8 0

2 years ago

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .