What is the formula for the surface area of a triangular pyramid

The length of a regulation soccer field is 110 meters. The diagonal length of the same soccer field is about 133.14 meters. Abou

dexar [7]

Answer:

D

Step-by-step explanation:

Imagine this as a right angle triangle, where the diagonal length is the hypotenuse, the length is one side, and the width is the other.

We can therefore use Pythagoras' Theorem (or Pythagorean Theorem) to solve. The formula for this is a²+b²=c², where c is the hypotenuse, and a and b are the sides.

We can input the values we know to this formula to get the width. This gives 110²+b²=133.14² or 12100+b²=17 726.2596.

From there subtracting 12100 from both sides gives b²=5626.2596.

Square rooting b isolates it, leaving b=75.0083969.

Since the value of the diagonal was approximate, this can be assumed the b is 75m.

**This content involves Pythagoras' Theorem/Pythagorean Theorem, which you may wish to revise. I'm always happy to help!

8 0

3 years ago

Muffins are packed and sold in boxes of 4. How many boxes are needed to pack 260 muffins?

Volgvan

Answer:

65 boxes

Step-by-step explanation:

Im not sure how to explain other then say all you do is divide 260 and 4

I hope this helped!

Good luck!

If i did help maybe you could help me by giving me brainliest!?

Thanks! <3

5 0

3 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

or

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .