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4 years ago

True or false:

Mathematics

1 answer:

Alinara [238K]4 years ago

8 0

1.true
2.true
3.because it supposed to be 4 sides like a rhombus
I’m not sure if they are right

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The temperature, T , of a given mass of gas varies inversely with its volume, V . The temperature of a certain gas with volume o

alekssr [168]

We know that

if t<span>he temperature T of a given mass of gas varies inversely with its volume V
</span>then
T=k/V

Step 1
Find the value of k
for T=30º C and V=105 cm³
we have
T=k/V--------> k=T*V--------> k=30*105=3150 °C*cm³

therefore
for V=84 cm³
T=3150/84=37.5 °C

the answer is 37.5 °C

3 0

3 years ago

In ΔQRS, the measure of ∠S=90°, the measure of ∠Q=29°, and SQ = 5 feet. Find the length of QR to the nearest tenth of a foot.

goldfiish [28.3K]

Answer:

2.8 feet

Step-by-step explanation:

tan(29)=x/5

5tan(29)=x

2.8=x

8 0

3 years ago

Read 2 more answers

I need the Combination( probability) of this problem.

coldgirl [10]

Check attached image file.

8 0

3 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution