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yaroslaw [1]
3 years ago
12

How do you write two hundred and three thousandths in number form

Mathematics
1 answer:
maks197457 [2]3 years ago
5 0
<span>
-- "Two hundred and three thousandths" means' ' 2.003 '.

-- "Two hundred three thousandths" would mean ' 0.203 '.

How can you tell the difference?

The difference is the 'and'.</span>  "And" means 'decimal point'.

' 678 ' is read as "six hundred seventy eight", with no 'and'.

' 40.06 ' is read as 'forty and six hundredths'.

' 0.46 ' is read as ' forty-six hundredths'.


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99 % of the trees in our neighborhood are eucalyptus trees. The town planning commission wants to get rid of some of these trees
Brut [27]

Answer:

50%

Step-by-step explanation:

Let the total number of trees be T and suppose that this is made up of eucalyptus trees E and other kinds of trees, O.

Then E+O=T, and since 99% of the trees are eucalyptus trees, EE+O=0.99.

Therefore

0.99E+0.99O=E

0.01E=0.99O

E=99O.

Suppose we reduce the number of eucalyptus trees by some amount x and that the remaining number of eucalyptus trees are now 98% of the total trees.

Then we would have E−x(E−x)+O=0.98.

Therefore,

0.02(E−x)=0.98O

0.02E−0.02x=0.98O

0.02x=0.02E−0.98O

x=E−50(0.98)O

x=E−49O

x=E−49(E99)

x=5099E≈0.505050...×E.

So we end up removing a little over half of the eucalyptus trees. EDIT: Forgot to mention - removing 50.505050...% of the eucalyptus trees means that we are removing 0.50505050....×0.99T which is 0.5T or half the trees in the neighborhood.

3 0
3 years ago
Decide whether each relation defines y as a function of x. Give the domain and range.
lisov135 [29]
Y=x².

This is te function of a parabola open upward

the domain is all values of x ={x/x∈z}
7 0
3 years ago
The altitude of a triangle is 1cm shorter than the base. If the area of the triangle is 15cm2, calculate the altitude.
WITCHER [35]

Answer:

5 cm

Step-by-step explanation:

We khow that the altitude of this triangle is 1cm shorter than the base

  • Let H be our altitude and B our base and A the area of the triangle
  • A= (B*H)/2 ⇒ 15=(B*H)/2
  • H is 1cm shorter than B ⇒ B=H+1
  • H*(H+1)/2=15 ⇒ H*(H+1)=30⇒ H²+H=30⇒H²+H-30+0

that's a quadratic equation . Let's calculate the dicriminant .

Let Δ be the dicriminant

  • a=1
  • b=1
  • c= -30
  • Δ=b²-4*a*c = 1²-4*1*(-30)=1+4*30=121≥0
  • Δ≥0⇔ that we have two solutions x and y
  • x= (-1-\sqrt{121})/2= (-1-11)/2= -6
  • y= (-1+\sqrt{121})/2= 10/2 = 5

We have a negative value and a positive one

The altitude is a distance so it can't be negative

H= 5cm

6 0
3 years ago
Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i
Leya [2.2K]
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.
 
Now, the system of equations in matrix form will be:
\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right]   \left[\begin{array}{ccc}x&\\y&\end{array}\right] =  \left[\begin{array}{ccc}375&\\7&\end{array}\right]

Next, we are going to find the determinant:
D=  \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}
Next, we are going to find the determinant of x:
D_{x} =  \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8

Now, we can find x:
x=  \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi

Now that we know the value of x, we can find y:
y=375-325=50mi

Remember that time equals distance over velocity; therefore, the time on the highway will be:
t_{h} = \frac{325}{65} =5hours
An the time on the city will be:
t_{c} = \frac{50}{25} =2hours

We can conclude that the bus was five hours on the highway and two hours in the city. 

8 0
3 years ago
Please help! Due soon! Graph the equation using the point and the slope.<br> y-3=1/5(x-1)
Nutka1998 [239]

Answer:

Check out the image below!

Step-by-step explanation:

3 0
3 years ago
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