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3 years ago

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

1 answer:

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:

Step-by-step explanation:

Your welcome

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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:
$-11=(-3)^2+7*(-3)+1$
$-11=9-21+1$
$-11=-11$

$79=6^2+7*6+1$
$79=36+42+1$
$79=79$

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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

280/56 = height

height = 5 meters

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 fourth 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 4th quadrant 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 .

Note: (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