If PS=16cm, RS=9cm and QS =12cm. What is the measurements of PQ and QR? Please help!!

Mathematics

1 answer:

Bond [772]2 years ago

5 0

Answer:

PQ = 28 cm

QR = 21 cm

Step-by-step explanation:

PQ = PS + SQ = 16 + 12= 28

QR = QS + SR = 12 + 9 = 21

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Enter a range of values for x.<br> 85°<br> 25<br> 5x - 10<br> [ ? ] Enter

Nataly_w [17]

<h3>Answer: 2 < x < 7</h3>

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Explanation:

We'll use the hinge theorem here. This says (more or less) that the larger an angle is, the side opposite of that will be longer.

Imagine that the segments with the single tickmarks represent a door swinging open/shut. The more open a door is, the further the distance it is from the handle to the frame. In terms of these triangles, the segment 25 is larger than 5x-10 because the angle 90 (opposite the 25) is larger than the angle 85 degrees (opposite the 5x-10)

In symbols we say

5x - 10 < 25

We also say 5x - 10 > 0 or 0 < 5x - 10 to ensure that the segment of length 5x-10 is not 0 or negative.

Put the two inequalities together and we get 0 < 5x - 10 < 25

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Solve for x

0 < 5x - 10 < 25

0+10 < 5x - 10 < 25 + 10 .... adding 10 to all sides

10 < 5x < 35

10/5 < 5x/5 < 35/5 ... dividing all sides by 5

2 < x < 7

We have x between 2 and 7, and not equal to either endpoint.

3 0

3 years ago

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .