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The shark are fed three times a day during the morning feeding 2/15 of a ton. During the afternoon feeding the weight of fish fe

UNO [17]

Answer:

The shark are fed $\frac16 \ ton$ of fish during the night.

Step-by-step explanation:

Given:

Weight of fish fed in the morning = $\frac{2}{15}\ ton$

Also Given:

During the afternoon feeding the weight of fish fed will be 1/15 of a ton more than the fish fed during the morning.

Weight of fish fed in the afternoon = $\frac{2}{15}+\frac{1}{15}=\frac{2+1}{15}=\frac{3}{15}\ ton$

Total fish fed in whole day = $\frac12 \ ton$

We need to find the Weight of fish fed in the night.

Solution:

Now we can say that;

Weight of fish fed in the night can be calculated by by subtracting Weight of fish fed in the morning and Weight of fish fed in the afternoon from Total fish fed in whole day .

framing in equation form we get;

Weight of fish fed in the night = $\frac{1}{2}-\frac{2}{15}-\frac{3}{15}= \frac{1}{2}-(\frac{2}{15}+\frac{3}{15})=\frac{1}{2}-\frac{2+3}{15}= \frac{1}{2}-\frac{5}{15} = \frac{1}{2}-\frac{1}{3}$

Now we will use LCM for making the denominator common we get;

Weight of fish fed in the night = $\frac{1\times3}{2\times3}-\frac{1\times2}{3\times2} = \frac{3}{6}-\frac{2}{6}$

Now denominator are common so we will solve the numerators.

Weight of fish fed in the night = $\frac{3-2}{6}=\frac16 \ ton$

Hence The shark are fed $\frac16 \ ton$ of fish during the night.

3 0

2 years ago

Use the following area model to multiply 42 x 73

Sergio [31]

A=2800,B=120,C=140,D=6

7 0

3 years ago

7- square root of 3i

r-ruslan [8.4K]

Step-by-step explanation:

Any complex number is of the form: $a + bi$ where a is real part and B is imaginary part.

Therefore, in given complex number: $7- \sqrt {3}i$

7 is real part.

$\sqrt {3}$ is imaginary part.

Hence, first two options are correct.

Last two options are incorrect because of the following reasons:

$- \sqrt {3}$ is coefficient of $i$ & not $7- \sqrt {3}$

Given number is a difference of a real number and imaginary number & not a sum of a real number and imaginary number.

5 0

3 years ago

Read 2 more answers

you have 3.75 to spend at a vending machine. You want to buy at least three health bars. Regualr health bars cost 0.75 each and

DedPeter [7]

Answer:

Let x be the number of regular health bars you buy and y the number of strawberry health bars you buy. Then:

0.75x+1.25y=3.75

x+y>=3

Step-by-step explanation:

For the first equation, we have to assume that you will spend all of your money, otherwise it becomes an inequation. The money you spend on regular bars is 0.75x dollars and the money you spend on strawberry bars is 1.25y, so if you spend your 3.75 dollars on the bars, then 0.75x+1.25y=3.75.

For the second, you will always buy x+y health bars, regular and strawberry. There isn't enough information to make this into a equation, the only thing we can deduce is the inequation x+y>=3.

If we also assume that x and y are integers (we can't buy half-bars or one-fourth of a bar) then the minimum number of bars we can buy is 3 (3 strawberry bars) and the maximum is 5 bars (5 regular bars). x+y must be an integer too, so the possibilities for the second equation are x+y=3, x+y=4 and x+y=5. There is a finite number of solutions in any case.

4 0

2 years ago

When solving a system of linear equations, try to algebraically form one equation that has only one variable.

lianna [129]

I believe the correct answer is true. <span>When solving a system of linear equations, try to algebraically form one equation that has only one variable. In this way, you can solve the value of that variable and eventually solve the other variables. Hope this answers the question. Have a nice day.</span>

3 0

3 years ago

Read 2 more answers