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3 years ago

PLEASE HELPPPPPPPPPPPPPPPPPPPPPP I BEG

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

Answer:

I think it's 4/10 which is 0.4

Step-by-step explanation:

bixtya [17]3 years ago

4 0

Answer:

I think the answer is 0.4 or 4/10 as well as the other guy

Step-by-step explanation:

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Rewrite the equation below so that it does not have fractions four -4/9 times equals two over three do not use decimals in your

Rama09 [41]

The equation is 36 - 4x = 6.

Solution:

To write the equation without fraction.

$$4-\frac{4}{9} x=\frac{2}{3}$

$$\frac{4}{1} -\frac{4}{9} x=\frac{2}{3}$

The denominators must be same to add/subtract the fractions.

LCM of 1, 9 and 3 = 9.

Multiply first term by $\frac{9}{9}$ and $\frac{2}{3}$ by $\frac{3}{3}$ to make the denominator same.

$$\frac{4\times9}{1\times9} -\frac{4}{9} x=\frac{2\times3}{3\times3}$

$$\frac{36}{9} -\frac{4}{9} x=\frac{6}{9}$

$$\frac{36-4x}{9} =\frac{6}{9}$

Multiply by 9 on both sides.

$$\frac{36-4x}{9}\times9 =\frac{6}{9}\times9$

Both 9 in the numerator and denominator get canceled.

36 - 4x = 6

This term does not have fraction.

Hence the equation is 36 - 4x = 6.

6 0

3 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,

$A=45e^{-0.0045(2500)}$ $A=45e^{-11.25}$ $A=45\times0.000013=0.000585$ $A=0.000585$

The amount of substance will be present around in 2500 years is 0.000585 grams

4 0

1 year ago

Find the area of the rhombus

professor190 [17]

Answer:

108

Step-by-step explanation:

none. I looked it up...so ion know lol. Im just tryna get these 50 points for sum reason...

3 0

3 years ago

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The heights of a maple tree and a cherry tree have a ratio of 5:2. If the maple tree grew 20 cm and 20 cm was cut off the top of

elena-14-01-66 [18.8K]

Let x be the <span>heights of a maple tree and y be the height of the cherry tree.
We know:
</span> $\frac{x}{y} = \frac{5}{2}$
<span>The new ratio is obtained like this:
</span> $\frac{x+20}{y-20} = \frac{3}{1}$ .
From the above equation we get $x= (y-20)\frac{3}{1} -20$ .
then $\frac{ (y-20)\frac{3}{1} -20}{y} = \frac{5}{2}$
Solving the above equation for y we get: $y=160$
So $x=400$
So the first tree is (400-160=240) more taller than cherry tree.

7 0

3 years ago

Which of the following values would be an outlier in the following set? 1, 14, 17, 18, 19, 23

tatuchka [14]

1 would be the outlier

5 0

3 years ago

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