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ss7ja [257]
2 years ago
13

(-8,2) and (3,5) slope

Mathematics
1 answer:
kvv77 [185]2 years ago
7 0

Answer:

Slope= 3/11

Step-by-step explanation:

Slope formula= \frac{y2-y1}{x2-x1}

Use the points (-8, 2) and (3,5)

\frac{5-2}{3-(-8)} = \frac{3}{3+8} = \frac{3}{11}

Slope= \frac{3}{11}

Use point-slope to find the equation

y-y1=m(x-x1)

*Use the slope (previously found) as m, and (3,5) as your point (x,y)*

y-5=\frac{3}{11}(x-3)

y-5=\frac{3}{11}x-3\frac{3}{11}

 +5      +5

y=\frac{3}{11}x+2\frac{3}{11}

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Molodets [167]

Answer:

x = 4√5

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Trigonometry</u>

  • Pythagorean Theorem: a² + b² = c²

Step-by-step explanation:

<u>Step 1: Define</u>

We are given a right triangle. We can use Pythagorean Theorem.

a = 19

b = x

c = 21

<u>Step 2: Find </u><em><u>x</u></em>

  1. Substitute:                    19² + x² = 21²
  2. Isolate <em>x</em> term:               x² = 21² - 19²
  3. Evaluate:                       x² = 441 - 361
  4. Subtract:                       x² = 80
  5. Isolate <em>x</em>:                       x = √80
  6. Simplify:                        x = 4√5

And we have our final answer!

8 0
3 years ago
Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°
natima [27]

 

\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0

.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

5 0
2 years ago
What is the slope of a line that passes through at (-1,5) and (4,5)
murzikaleks [220]
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3 0
3 years ago
Whats the length of A, B and C
kondaur [170]

Answer:

B should be 3 inches long because the area of the square above is 48 inches squared, so that long side is probably 8. B is 3 inches long. As for A, this should be simple. since we know the longer side of the room has the area of 27, we can tell the length(longest side) is 9. 14 minus 9 is equal to 4. A's width is 4 inches long.

Step-by-step explanation:

for the rest, since we know one of the widths is 6 inches, we know that the length of the room with an area of 32 inches is 8. therefore, since 8 times 4 equals the number 32, C is 4 inches long. back to B. since we know that c is 4 inches long and that there is a length of 11 inches, we can tell the longest side of room A is 7. 4 Times 7 equals 28, so B is 28 inches squared.

3 0
3 years ago
PLEASE HELP ASAP!!!
EleoNora [17]
<span>Part 1:

(x+1)^2-3(x+2) \\  \\ =x^2+2x+1-3x-6 \\  \\ =x^2-x-5

option b.


Part 2:
(5x^3y^{-3})(-6x^4y)\\ \\=5(-6)x^{3+4}y^{-3+1}\\ \\=-30x^7y^{-2}\\ \\=\frac{-30x^7}{y^2}

option d


Part 3:
\frac{8+x}{2} - \frac{x+3}{3} =4

Multiply through by the LCM of 2 and 3 (i.e. 6)
6\left(\frac{8+x}{2}\right) - 6\left(\frac{x+3}{3}\right) =6(4) \\  \\ 3(8+x)-2(x+3)=24 \\  \\ 24+3x-2x-6=24 \\  \\ x+18=24 \\  \\ x=24-18=6

option b


Part 4
x^2-3x-8=x+4

x^2-3x-8=x+4 \\  \\ x^2-3x-x-8-4=0 \\  \\ x^2-4x-12=0 \\  \\ (x-6)(x+2)=0 \\  \\ x=6 \ or \ -2

option d


Part 5
Joseph is solving the equation x^2+8x-4=0 using the technique completing the square.
x^2+8x-4=0
x^2+8x+?=4+?

In completeing the square method, you divide the coeficient of x by 2, square the result and add to both sides of the equation.

Therefore, he should add 16 to both sides because (8/2)^2=16.

option d.
</span>
5 0
3 years ago
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