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2 years ago

Round the quantity to th given place value 13.989 to the nearest tenth

Mathematics

2 answers:

MA_775_DIABLO [31]2 years ago

5 0

Answer:

Step-by-step explanation:

round

Aliun [14]2 years ago

3 0

Answer:

13.989 rounded to nearest tenth is 14.0.

Step-by-step explanation:

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There are 12 boys and 15 girls in your class for new students three boys and one girl join your class what is the ratio of the c

Finger [1]

Before the new kids the ratio of girls to boys is 15:12 because there are 15 girls and 12 boys now you can simplify by dividing 15 by 3 and 12 by 3 to get 5:4 as your simplified ratio. As far as after the new kids. Since there is 1 new girl you do 15+1 to get 16 and there are 3 new boys so you do 12+3 to get 15 now 16:15 is already in simplest form so your answer is:

Before: 15:12 simplified as 5:4
After 16:15

6 0

3 years ago

Read 2 more answers

Please help show work<br><br> 3√(125) -√(16) = ?

Anna007 [38]

First, simplify the radicals
√125=√(5*5*5)=(√5*5)(√5)=5√5
so 3√125=3*5√5=15√5

and √16=4
so
15√(5)-4 is the simplified version

5 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$

6 0

3 years ago

-9f +19-1 + f) 12-=8f <br>is this the answer?

jeka94

Hi!

-9f + 19 - 1 + f is actually -8f + 18.

To do this we add like terms, -9f and f; and 19 and -1

-9f + f is -8f

19 - 1 is 18

Therefore your answer is -8f + 18

7 0

2 years ago

The ages of undergraduate students in a state

Elan Coil [88]

The percent of students that are aged 19 years or more is determined as 84%.

<h3>One standard deviation below the mean</h3>

In a normal distribution curve 1 standard deviation below the mean is defined as follows;

1 std below mean : M - d = 16%

M - d = 20.6 yrs - 1.3 yrs = 19.3 years ≈ 19 years

19 years or more will occur at (M - d) + (M) + (M + 2d) = 100% - (M - d)

= 100% - 16%

= 84%

Thus, the percent of students that are aged 19 years or more is determined as 84%.

Learn more about normal distribution here: brainly.com/question/4079902

#SPJ1

7 0

2 years ago