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ololo11 [35]
3 years ago
14

What is the distance between the two points (5,-2) and (-3,8)

Mathematics
1 answer:
bearhunter [10]3 years ago
4 0
You need to use the distance formula
d =  \sqrt{ {(x - x)}^{2}  +  {(y - y)}^{2} }

\sqrt{ {(5 + 3)}^{2}  +  {( - 2 - 8)}^{2} }
so the distance between points (5,-2) and (-3,8) is
2 \sqrt{41}
which won't simplify so it stays as is
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If f(x) = 0, what is x?
joja [24]

Answer:

0

Step-by-step explanation:

Your welcome

5 0
3 years ago
Which quadratic equation fits the data in the table? (PICTURE OF TABLE ATTACHED)
Paha777 [63]
The correct quadratic equation for that set of points is:
y=x^2+7x+1

To check it, simply put first and last pair of points and assure the equality:
-11=(-3)^2+7*(-3)+1
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-11=-11

79=6^2+7*6+1
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5 0
3 years ago
What is the height of a rectangular prism that has a volume of 280 cubic meters,a length of 8 meters, and a width of 7 meters? S
lyudmila [28]
When looking at a rectangular prism, we know that Volume = length * width * height

If we are given that the volume of the prism is 280 cubic meters (m^3), the length is 8 meters, and the width is 7 meters, we can fill in our equation.

280 = 8 * 7 * height
280 = 56 * height

Because we are solving for the height, we divide both sides by 56

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8 0
3 years ago
The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis
Firdavs [7]

Answer:

sin\theta_1 =  - \frac{2\sqrt{70}}{19}

Step-by-step explanation:

We are given that \theta_1 is in <em>fourth</em> quadrant.

cos\theta_1 is always positive in 4th quadrant and  

sin\theta_1 is always negative in 4th quadrant.

Also, we know the following identity about sin\theta and cos\theta:

sin^2\theta + cos^2\theta = 1

Using \theta_1 in place of \theta:

sin^2\theta_1 + cos^2\theta_1 = 1

We are given that cos\theta_1 = \frac{9}{19}

\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{280}{361}\\\Rightarrow sin\theta_1 =  \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 =  +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}

\theta_1 is in <em>4th quadrant </em>so sin\theta_1 is negative.

So, value of sin\theta_1 =  - \frac{2\sqrt{70}}{19}

6 0
3 years ago
Write the quadratic equation in general form. (x - 3)^ 2 = 0
algol [13]
The answer is:  x² – 6x + 9 = 0  .
_____________________________________________________
Explanation:
________________________________________________________

Given:  (x – 3)² = 0 ;  write as: general form:  "ax² + bx + c = 0";  a ≠ 0 .
<span>
Note:  </span>(x – 3)² = (x – 3)(x – 3) = x² – 3x – 3x + 9 = x² – 6x + 9 ;
___________________________________________________
Rewrite: (x – 3)²  =  0 ;  →
           as:   x² – 6x + 9 = 0  ; which is our answer.
____________________________________________
→   x² – 6x + 9 = 0  ; is in "general form", or "standard equation format"; that is: " ax² + bx + c = 0 ";  (a ≠ 0) ;
             → in which: 
   a =  1 (implied coefficient, since anything multiplied by "1" is that same                                                                                                                    value);
   b = -6; 
   c =  9
_______________________________________________________
5 0
3 years ago
Read 2 more answers
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