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3 years ago

From the equation y−3=−2(x+1), identify the slope and a point on the line.

Mathematics

1 answer:

olga2289 [7]3 years ago

4 0

Answer:

slope = - 2, point on line = (- 1, 3 )

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

y - 3 = - 2(x + 1) ← is in point- slope form

with slope m = - 2 and (a, b ) = (- 1, 3 )

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Zack, Rachel, and Maddie are unraveling a huge ball of yarn to see how long it is. As they move away from each other, they form

Korolek [52]

We are tasked to solve the three angles given that we have the measurements of the sides such as
a = Zack to Rachel distance
b = Rachel to Maddie distance
c = Maddie to Zack distance

a =3
b =2.5
c =4

Solving the angles we need to use Law of Cosines:
cos A = 2.5² + 4² -3² /2*2.5*4
A = 48.59°

cos B=3² + 4² - 2.5² / 2*3*4
B = 38.625°

C=180 - 48.59° - 38.625°
C= 92.79°

The three angles are 48.59°,38.63° and 92.79°.

8 0

3 years ago

Can some one help me

il63 [147K]

Answer:

5/6

Step-by-step explanation:

Dividing fractions:

Step 1: Rewrite the first fraction as it is.

Step 2: Replace the division sign with a multiplication sign.

Step 3: Flip the second fraction.

Step 4: Multiply the fractions and reduce the product if necessary.

Let's use the rule of dividing fractions on your problem.

Step 1: Rewrite the first fraction as it is.

$\dfrac{5}{8}$

Step 2: Replace the division sign with a multiplication sign.

$\dfrac{5}{8} \times$

Step 3: Flip the second fraction.

$\dfrac{5}{8} \times \dfrac{4}{3}$

Step 4: Multiply the fractions and reduce the product if necessary.

To multiply fractions, multiply the numerators together, and multiply the denominators together.

$\dfrac{5}{8} \times \dfrac{4}{3} = \dfrac{5 \times 4}{8 \times 3} = \dfrac{20}{24}$

We notice that the greatest common factor of 20 and 24 is 4, so we divide both the numerator and denominator by 4 to reduce the fraction.

$= \dfrac{4 \times 5}{4 \times 6} = \dfrac{5}{6}$

5 0

3 years ago

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$