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3 years ago

A line passes through (2, −1) and (4, 5).

Mathematics

1 answer:

viktelen [127]3 years ago

5 0

the answer to your question is
−3x+y=−7

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Find the missing part.

Katarina [22]

Answer:

$z = 15[\frac{\sqrt{3}}{2}]$

Step-by-step explanation:

To find the Z-side, we must take the cosine of the 30-degree angle of the main triangle

We know that the cosine of an angle is defined as:

$cos(30) = \frac{adjacent\ side}{hypotenuse}$

$cos(30) = \frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}} = \frac{z}{15}$

Then:

$z = 15[\frac{\sqrt{3}}{2}}]$

Finalmente the side z is:

$z = 15[\frac{\sqrt{3}}{2}}]$

7 0

4 years ago

Read 2 more answers

Please help me yhank you

Citrus2011 [14]

Answer:972

Step-by-step explanation:

Divide both numbers by 972 and they both go into each other

5 0

4 years ago

What is the equation for the line of reflection? On a coordinate plane, triangle A B C has points (6, 3.7), (5.4, 2), (1, 3). Tr

Klio2033 [76]

Answer:

The equation of the line of reflection will be x = y.

Step-by-step explanation:

On a coordinate plane, triangle Δ ABC has points (6, 3.7), (5.4, 2), (1, 3). And triangle Δ A'B'C' has points (3.7, 6), (2, 5.4), (3, 1).

Now, Δ A'B'C' is the image triangle of Δ ABC and we have to find the equation of the line of reflection.

We know that after reflection of a point P(h,k) over the line x = y the image will be P'(k,h).

Same is happened in case of Δ ABC and Δ A'B'C'.

Therefore, the equation of the line of reflection will be x = y. (Answer)

3 0

4 years ago

Let z = ln(x 2 + y), x = ret . and y = ter . Use the Chain Rule to compute ∂z ∂r and ∂z ∂t at the point where (r, t) = (1, 2).\

Natali [406]

By the chain rule,

$\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}$

where $u\in\{r,t\}$ .

We have component partial derivatives

$\dfrac{\partial z}{\partial x}=\dfrac{2x}{x^2+y}=\dfrac{2re^t}{r^2e^{2t}+te^r}$

$\dfrac{\partial z}{\partial y}=\dfrac1{x^2+y}=\dfrac1{r^2e^{2t}+te^r}$

$\dfrac{\partial x}{\partial r}=e^t$

$\dfrac{\partial x}{\partial t}=re^t$

$\dfrac{\partial y}{\partial r}=te^r$

$\dfrac{\partial y}{\partial t}=e^r$

Putting the appropriate pieces together and setting $(r,t)=(1,2)$ , we get

$\dfrac{\partial z}{\partial r}(1,2)=\dfrac{2e^3+2}{e^3+2}$

$\dfrac{\partial z}{\partial t}(1,2)=\dfrac{2e^3+1}{e^3+2}$

3 0

3 years ago

Simplify<br> 1. √28<br> 2. √72<br> 3. √2 times √10<br> Please help

Mnenie [13.5K]

Answer:

Step-by-step explanation:

1. 2 square root 7

2. 6 square root 2

3. 2 square root 5

5 0

3 years ago