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GREYUIT [131]
3 years ago
13

7 3/4 write as a decimal

Mathematics
2 answers:
Ad libitum [116K]3 years ago
7 0

Answer:

7.75

mark me brainlest

Step-by-step explanation:

irina [24]3 years ago
5 0

Answer:

7.75

Step-by-step explanation:

3/4 = 0.75

think about it as how many quarters make a dollar, 4, so 3 quarters makes 75 cents (only works with fractions with a denominator of 4).

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A bus is travelling from Addis Ababa to Ambo. It travels 43 km in the first hour
zhuklara [117]
<h3>Answer:  43 km per hour</h3>

==================================================

Work Shown:

total distance = 43+40+46 = 129 km

total time = 1+1+1 = 3 hours

average speed = (total distance)/(total time)

average speed = (129 km)/(3 hrs)

average speed = (129/3) (km/hr)

average speed = 43 km per hour

8 0
2 years ago
Read 2 more answers
Without solving determine the number of real solutions for each quadratic equation
seropon [69]

Answer:

b^2-4b+3=0

b²-3x-b+3=0

b(b-3)-1(b-3)=0

(b-3)(b-1)=0

either

b=3 or b=1

.

2n^2 + 7 = -4n + 5

2n²+4n+7-5=0

2n²+4n+2=0

2(n²+2n+1)=0

(n+1)²=0/2

:.n=-1

.

x - 3x^2 = 5+ 2x - x^2

0=5+ 2x - x^2-x +3x^2

0=5+x+2x²

2x²+x+5=0

comparing above equation with ax²+bx +c we get

a=2

b=1

c=5

x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1

={-1±√-39}/2

3 0
2 years ago
A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant prod
-Dominant- [34]

Answer:

Step-by-step explanation:

Given that a farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces

When we consider this graph as a straight line, the two points lying on the line would be

(30, 30) and (34, 28) taking n as horizontal and y vertical

Using two point equation we find that

the equation of the line is

\frac{y-y_1}{y_2-y_1} =\frac{x-x_1}{x_2-x_1}

Substitute the points as x =n

\frac{y-30}{28-30} =\frac{x-30}{34-30}\\y-30 = -0.5(x-30)\\y = -0.5x+45

is the linear relationship between n and y

7 0
2 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
Write any 4 laws of exponents.<br>​
emmainna [20.7K]

         \rule{50}{1}\large\blue\textsf{\textbf{\underline{Question:-}}}\rule{50}{1}

       <em>Write any 4 laws, or properties, of exponents.</em>

<em />

<em>        </em>\rule{50}{1}\large\blue\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}

\Large\text{Law number 1:-}

\Large\text{$a^n*a^m=a^{n+m}$}

\Large\text{It states that:-}

                        When we multiply together numbers with the same base,

                         we add the exponents.

\Large\text{Law number 2:-}

\Large\text{$a^m:a^n=a^{m-n}$}

\Large\text{It states that:-}

                     When we divide numbers with the same base, we

                     subtract the exponents.

\Large\text{Law number 3:-}

\Large\text{$(\displaystyle\frac{x}{y}) ^n=\frac{x^n}{y^n}$}

\Large\text{It states that:-}

                 If we have a fraction to a power, we raise the numerator and

               the denominator to that power.

And then last but not least,

\Large\textit{Law number 4:-}

\Large\text{$a^{-m} =\displaystyle\frac{1}{a^m}$}

\Large\text{It states that:-}

                   If we have a number with a negative exponent, we flip it over.

<h3>Good luck with your studies.</h3>

#TogetherWeGoFar

\rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}

6 0
2 years ago
Read 2 more answers
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