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2 years ago

Tom bought a coat for £49 in a 30% off sale. How much money did he save?

Mathematics

2 answers:

motikmotik2 years ago

7 0

Answer:

£13 is what time saved . thank you

Serjik [45]2 years ago

6 0

Answer:

21 pounds

Step-by-step explanation:

30% off sale means the price was reduced to 70%

70% of the initial price is 49pounds

Initial price = x

0.7x = 49

x = 49/.07 = 70

Amount saved = 70 - 49 = 21 pounds

Please mark brainliest if helpful. Thanks

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An object with a mass of 3.0 g raises the level of water in a graduated cylinder from 25.1 mL to 33.1 mL. What is the density of

evablogger [386]

The displaced volume of the water, because of the submersion of the object, is equal to the volume of the object.

So to find the volume of the object we calculate the difference in volume of water:
33.1-25.1 = 8 ml

Since 1 mL of water has the volume of 1cm^3, then the volume of the object is 8 cm^3.

The density of the object is found by the formula:

density= mass/volume= (3 g) / (8 cm^3) = 3/8 (g/cm^3)= 0.375 g/cm^3

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2 years ago

In Exercises 40-43, for what value(s) of k, if any, will the systems have (a) no solution, (b) a unique solution, and (c) infini

svet-max [94.6K]

Answer:

If k = −1 then the system has no solutions.

If k = 2 then the system has infinitely many solutions.

The system cannot have unique solution.

Step-by-step explanation:

We have the following system of equations

$x - 2y +3z = 2\\x + y + z = k\\2x - y + 4z = k^2$

The augmented matrix is

$\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]$

The reduction of this matrix to row-echelon form is outlined below.

$R_2\rightarrow R_2-R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]$

$R_3\rightarrow R_3-2R_1$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]$

$R_3\rightarrow R_3-R_2$

$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

$k^2-k-2=0$

$\left(k+1\right)\left(k-2\right)=0\\k=-1,\:k=2$

Case k = −1:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]$

If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$