All categories

3 years ago

What is 0.9 as a fraction

Mathematics

2 answers:

Vinvika [58]3 years ago

8 0

Using the place value chart, we can see that the decimal 0.9 is nine tenths, so we can write 0.9 as the fraction 9 out of 10 or $\frac{9}{10}$ .

The fraction is in lowest terms.

Therefore, 0.9 can be written as the fraction $\frac{9}{10}$ , which is in lowest terms.

Image is provided.

Allushta [10]3 years ago

3 0

Answer:

9/10

Step-by-step explanation:

You can write 9 as the numerator (top number), and 10 as the denominator (bottom number), and that will be equivalent to 0.9

You might be interested in

Please help i will give brainliest

Pepsi [2]

Answer:

Step-by-step explanation:

1000m=1k

3,500m=3.5k

500m=.5k

75m=0.075k

1m=0.001

7 0

3 years ago

Read 2 more answers

Identify the usual level of measurement for each of the following A. year in school B. IQ scores C. life expectancy D. fatigue E

Mashutka [201]

Answer:

The answer is C

Step-by-step explanation:

4 0

3 years ago

Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.