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3 years ago

Write an equation in standard form for each line.

Mathematics

1 answer:

LenaWriter [7]3 years ago

7 0

The standard form: Ax + By = C.

The point-slope form:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

7. We have the points (4, -7) and (2, -3). Substitute:

$m=\dfrac{-3-(-7)}{2-4}=\dfrac{4}{-2}=-2\\\\y-(-7)=-2(x-4)\qquad|\text{use distributive property}\\\\y+7=(-2)(x)+(-2)(-4)\\\\y+7=-2x+8\qquad|-7\\\\y=-2x+1\qquad|+2x\\\\\boxed{2x+y=1}$

8. We have the points (1, 5) and (-10, -6). Substitute:

$m=\dfrac{-6-5}{-10-1}=\dfrac{-11}{-11}=1\\\\y-5=1(x-1)\\\\y-5=x-1\qquad|+5\\\\y=x+4\qquad|-x\\\\-x+y=4\qquad|\text{change the signs}\\\\\boxed{x-y=-4}$

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Please help due today!

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Step-by-step explanation:

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3 years ago

Which is equivalent to (3 + 2i)(5 + i)?

zepelin [54]

(3 + 2i)(5 + i)
15 + 3i + 10i + 2i^2
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5 0

3 years ago

For questions 3 – 5, an airplane is heading south at an airspeed of 540 km/hr, but there is a wind blowing from the northeast at

GuDViN [60]

First, let's calculate the horizontal and vertical components of the wind speed (W) and the airplane speed (A), knowing that south is a bearing of 270° and northeast is a bearing of 45°:

$\begin{gathered} W_x=W\cos45°\\ \\ W_x=50\cdot0.707\\ \\ W_x=35.35\\ \\ \\ \\ W_y=W\sin45°\\ \\ W_y=50\cdot0.707\\ \\ W_y=35.35 \end{gathered}$ $\begin{gathered} A_x=A\cos270°\\ \\ A_x=540\cdot0\\ \\ A_x=0\\ \\ \\ \\ A_y=A\sin270°\\ \\ A_y=540\cdot(-1)\\ \\ A_y=-540 \end{gathered}$

Now, let's add the components of the same direction:

$\begin{gathered} V_x=W_x+A_x=35.35+0=35.35\\ \\ V_y=W_y+A_y=35.35-540=-504.65 \end{gathered}$

To find the resultant bearing (theta), we can use the formula below:

$\begin{gathered} \theta=\tan^{-1}(\frac{V_y}{V_x})\\ \\ \theta=\tan^{-1}(\frac{-504.65}{35.35})\\ \\ \theta=-86° \end{gathered}$

The angle -86° is equivalent to -86 + 360 = 274°.

Therefore the correct option is b.

8 0

1 year ago

What is 2.9 as a mixed number?

lapo4ka [179]

Answer:

2910 ….....................….......

3 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST! NEED QUICK HELP ASAP! Photo added! least common denominator x^3/x^2+4x+4 and -9/x^2+5x+6

Oksanka [162]

The denominator( s ) we are given are $x^2 + 4x + 4$ , and . The first thing we want to do is factor the expressions, to make this easier -

$x^2 + 4x + 4$

This expression is a perfect square, as ( x )^2 = x^2, ( 2 )^2 = 4, 2 * ( x ) * ( 2 ) = 4x. Thus, the simplified expression should be the following -

$( x + 2 )^2$

The other expression is, on the other hand, not a perfect square so we must break this expression into groups and attempt factorization -

$x^2 + 5x + 6,\\\left(x^2+2x\right)+\left(3x+6\right),\\x\left(x+2\right)+3\left(x+2\right),\\\left(x+2\right)\left(x+3\right)\\\\Factored Expression = \left(x+2\right)\left(x+3\right)$

Combining ( x + 2 )^2 and ( x + 2 )( x + 3 ), the expression that contains factors of each is ( x + 2 )^2 * ( x + 3 ), or in other words the LCM.

Solution =  $\left(x+2\right)^2\left(x+3\right)$ 

3 0

3 years ago