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3 years ago

A^2 + b^2 = c^2 (solve for c)

Mathematics

1 answer:

saul85 [17]3 years ago

8 0

c = √a^2+b^2

c = -√a^2+b^2

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The temperature of a liquid is decreasing at a constant rate. After two minutes of cooling it was 178 degrees and after 5 minute

enot [183]

Answer:

Step-by-step explanation:

If the liquid is cooling at a constant rate, then the equation for this is linear. We can use the info given to find the linear equation for this situation. We have 2 points: (2, 178) and (5, 154), where the x coordinate represents the time in minutes, and y represents the temperature. We use these values in the slope formula:

$m=\frac{154-178}{5-2}=-\frac{24}{3}=-8$

Now we will use one of the points to finish writing the equation. I chose (2, 178), but it doesn't matter which one you choose; they will both give you the same equation in the end.

y - 178 = -8(x - 2) and

y - 178 = -8x + 16 so

y = -8x + 196

If we are looking for the temp when the liquid starting cooling, then we are looking for the temp at time 0, or when x = 0 (sincce x is our time).

y = -8(0) + 196 so

y = 196

That's the temp of the liquid before it started any cooling.

3 0

4 years ago

G is the centroid of AABC, FC = 35, AG = 42, BF = 57, and DG = 14. Find BG.

Verizon [17]

Answer:

$BG = 38$

Step-by-step explanation:

Based on centroid theorem of triangles:

$BG = \frac{2}{3} \times BF$

$BF = 57$

Plug in the value into the equation

$BG = \frac{2}{3} \times 57$

$BG = \frac{2 \times 57}{3}$

$BG = \frac{114}{3}$

$BG = 38$

5 0

3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

Mandarinka [93]

Answer:

The factors of $x^{3}-y^{3}$ are $[(x-y)(x^{2}+xy+y^{2})]$ .

Step-by-step explanation:

We know that:

$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$

Compute the factors of $x^{3}-y^{3}$ as follows:

$x^{3}+(-y)^{3}=(x+(-y))(x^{2}-x(-y)+(-y)^{2})$

$x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$

Thus, the factors of $x^{3}-y^{3}$ are $[(x-y)(x^{2}+xy+y^{2})]$ .

3 0

3 years ago

If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta

Ivahew [28]

Step-by-step explanation:

We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

$\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}$

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

PQ = Opposite side
QR = Adjacent side
RP = Hypotenuse
∠Q = 90°
∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

$\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}$

Since, we know that,

cosθ = 5/13
QR (Adjacent side) = 5
RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

$\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}$

$\rule{200}{3}$

$\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

$\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}$

Where,

Opposite side = 12
Hypotenuse = 13

Therefore,

$\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}$

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

$\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}$

By substituting the values, we get,

$\rule{200}{3}$

$\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the required answer is 17/7.

6 0

3 years ago

What’s 9+10<br><br> A:21<br> B:19<br> C:12<br> D:21

krek1111 [17]

Answer:

Step-by-step explanation:

10 + 9 = 19

7 0

3 years ago