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algol13
3 years ago
5

After six weeks the tomato plant that was given extra food and water was 26 cm tall . The tomato plant that was not given any ex

tra food was only 74.5% as tall . How tall was the tomato plant that was no given extra plant food?
Mathematics
1 answer:
Len [333]3 years ago
5 0
The answer is 19.37 cm. The problem is only asking what's 74.5% of 26 cm, 0.745 • 26 = 19.37 cm.
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An aquarium measures 61 cm long, 30.5 cm wide, and 30.5 cm high.If the tank is only filled to 80% capacity with water, how much
LiRa [457]

Answer:

V=45396.2\ cm^3

Step-by-step explanation:

Given that,

The dimensions of an aquarium are 61 cm long, 30.5 cm wide, and 30.5 cm high.

The tank is filled to 80% capacity with water.

We need to find the water needed for this tank.

The volume of cuboid is given by :

V=lbh

As it is filled to 80% capacity, so,

V=\dfrac{80}{100}\times lbh\\\\=0.8\times 61\times 30.5\times 30.5\\\\V=45396.2\ cm^3

So, 45396.2\ cm^3 of water is needed for this tank.

5 0
2 years ago
Just number 2 please
Nezavi [6.7K]

Answer:

It has 3 sides and the measure is 3

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Solve for x please help ! (show work)
Alina [70]

Answer:

x = -5

Step-by-step explanation:

-(5x-2) = 27

Distribute the minus sign

-5x +2 = 27

Subtract 2 from each side

-5x +2-2 = 27-2

-5x = 25

Divide by -5

-5x/-5 = 25/-5

x = -5

5 0
3 years ago
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Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -6x - 8y = -2
evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}

Step # 1 : Swap the first and second matrix rows,

\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}

Step # 2 : Cancel leading coefficient in row 2 through R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1,

\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0
3 years ago
The line with equation y + 2x = 0 coincides with the terminal side of an angle θ in standard position in Quadrant IV .
Ket [755]

Answer:

-2

Step-by-step explanation:


7 0
3 years ago
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