All categories

2 years ago

A triangle has a side lengths of 4cm,3cm, and 4cm. Classify its as acute, obtuse, or right.

Mathematics

1 answer:

DIA [1.3K]2 years ago

4 0

When given 3 triangle sides, to determine if the triangle is acute, right or obtuse:

1) Square all 3 sides.

4, 3, 4,

16, 9, 16

2) Sum the squares of the 2 shortest sides.

16 + 9 = 25

3) Compare this sum to the square of the 3rd side.

25 > 16

if sum > 3rd side² Acute Triangle

So, it is an acute triangle.

Source:
http://www.1728.org/triantest.htm

You might be interested in

Which of the following shows the polynomial below written in descending order?

Juli2301 [7.4K]

Answer:

Step-by-step explanation:

The values must decrease
x∧12≥3x∧7≥4x³≥9x

4 0

2 years ago

Tell two different ways to find the difference of 414 subtract 178

sesenic [268]

178 + 336 = 414 and done

4 0

2 years ago

A student is solving the equation 4^x–1 = 64^x+3.

Doss [256]

The answer is x - 1 = 3x + 9

All steps are:
Step 1: 4ˣ⁻¹ = 64ˣ⁺³
Step 2: 4ˣ⁻¹ = (4³)ˣ⁺³ 
Step 3: 4ˣ⁻¹ = 4³ˣ⁺⁹
Step 4: x - 1 = 3x + 9
Step 5: x - 3x = 1 + 9
Step 6: -2x = 10
Step 7: x = 10/-2
Step 8: x = -5

4 0

3 years ago

Read 2 more answers

Add: (–21) + (–37) + (–15)

Rina8888 [55]

Answer:

the answer is -73

Step-by-step explanation:

4 0

2 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:

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