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3 years ago

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write an equation in standard form for a line that is (a) parallel (b) to the line with an equation of y=3x-5 that passes throug

h the point (8,5)

1 answer:

____ [38]3 years ago

4 0

Answer:

The parallel line would be y = 3x - 19

Step-by-step explanation:

In order to find the equation of this line, we first have to note that parallel lines have same slopes. Therefore, since the original line has a slope of 3, we know the new line will have a slope of 3. Now we can use that information along with the given point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - 5 = 3(x - 8)

y - 5 = 3x - 24

y = 3x - 19

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Using the quadratic formula, what are the solutions to x2 – 3x = -2?

Kryger [21]

Step-by-step explanation:

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6 0

2 years ago

Read 2 more answers

In Haley's favorite video game, players stack blocks to create buildings. Haley builds a fort shaped like a cube. Haley's fort h

Mama L [17]

Answer:

729 cubic meters

Step-by-step explanation:

Volume of cube is given by side^3

where side is length of the side

Given in the problem

side is 9 meters as length, width, and height of fort is 9 meters and it is in cube shape .

Thus,

volume of cube = 9^3 = 9*9*9 = 729

Thus, volume of Haley's fort is 729 cubic meters.

3 0

3 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

4 0

3 years ago

George observes that for every increase of 1 in the value of x, there is a increase of 60 in the corresponding value of y. He cl

Brums [2.3K]

Hi there!

<u><em>FACTS</em></u><em> :</em>

<em>To see if multiple ratios are proportional, you can write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportionnal ratios. You can also write them as fractions and divide the numerator (top number) by its denominator (bottom number), and compare the decimal numbers the same way you would compare the fractions (I personnaly find this method easier because you don't need to simplify the fractions).</em>

<u>STEPS TO ANSWER:</u>

x = 1 ; y = 90 → $\frac{1}{90} =$ 1 ÷ 90 = <u>0.01111111...</u>

x = 2 ; y = 150 → $\frac{2}{150} =$ 2 ÷ 150 = <u>0.0133333...</u>

x = 3 ; y = 210 → $\frac{3}{210} =$ 3 ÷ 210 = <u>0.01428571</u>

x = 4 ; y = 270 → $\frac{4}{270} =$ 4 ÷ 270 = <u>0.0148148...</u>

x = 5 ; y = 330 → $\frac{5}{330} =$ 5 ÷ 330 = <u>0.01515152</u>

<em>** You didn't really need to calculate them all because even the first two decimal numbers weren't equivalent, but I wanted to show you the whole process so I calculated them all.</em>

⇒ If you compare all the decimals you got, you can easily see that they are not the same, which means that the ratios between the values of "x" and the values of "y" are not proportional. Therefore, George's reasoning wasn't good.

There you go! I really hope this helped, if there's anything just let me know! :)

5 0

4 years ago

use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle

Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

$\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

$\int_C9y^3dx-9x^3dy$

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

$\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2$

$\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2$

$\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy$

⇒ $-27\int\int_D(x^2 + y^2)dx dy$

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

$-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta$

⇒ $-27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta$

⇒ $-27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta$

⇒ $-27\int\limits^{2\pi}_0 (16/4) d\theta$

⇒ $-108\int\limits^{2\pi}_0 d\theta$

⇒ $-108[2\pi - 0]$

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

#SPJ4

3 0

2 years ago