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3 years ago

Between which 2 integers does the negative square root of 33 lie?

Mathematics

2 answers:

Alexxandr [17]3 years ago

5 0

the answer would be letter E: -5 and -6

Step-by-step explanation:

igomit [66]3 years ago

4 0

Umm I think it’s b -6 and -7 or -5 and -6

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Your international diplomacy trip requires stops in thailand, singapore, hong kong, laos, and bali. how many possible itinerarie

Stells [14]

I would say there would be 5 itineraries ie one each for Thailand, Singapore, Hong Kong, Laos and Bali though for the countries Thailand and Laos there may be more than one stop so there would be a minimum of 5 itineraries.

6 0

4 years ago

a trapezoid has base lengths of (6x-1) units and 3 units. Its midsegment has a length of (5x-3) units. What is the value of x?

vodka [1.7K]

1. You have that:

- The<span> lengths of the bases are (6x-1) units and 3 units.
- The midsegment has a length of (5x-3) units.

2. To solve this exercise, you must apply the formula for calculate the length of the midsegment of a trapezoid, which is shown below:

Midsegment=Base1+Base2/2

As you can see, the midsegment is half the sum of the bases of the trapezoid.

3. When you substitute the values, you obtain:

(5x-3)=[(6x-1)+3]/2

4. Now, you can solve the problem by clearing the "x":
</span>
(5x-3)=[(6x-1)+3]/2
2(5x-3)=6x-1+3
10x-6=6x+2
10x-6x=2+6
4x=8
x=8/4
x=2

3 0

3 years ago

Hola pueden ayudar a=6cm b=8cm c?

Alexxandr [17]

Answer:

c=10cm

Step-by-step explanation:

a²+b²=c²

6²+8²=c²

36+64=c²

100=c²

10=c

6 0

2 years ago

NO LINKS!!! Please help me with these problems

nadezda [96]

<h3>Answers:</h3>

7) Center= (-1,2) Radius= $\boldsymbol{\sqrt{8}}$ Equation: $(x+1)^2+(y-2)^2 = 8$

8) Center= (3,13) Radius= 13 Equation: $(x-3)^2+(y-13)^2 = 169$

=========================================================

Explanation:

Problem 7

Let's find the distance from (-1,2) to (-3,4)

$(x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\$

This is the radius because it stretches from the center to a point on the circle, so $r = \sqrt{8}$

Squaring both sides will get us $r^2 = 8$

One useful template for a circle is the equation $(x-h)^2+(y-k)^2 = r^2\\\\$

(h,k) is the center

r is the radius

Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.

$(x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\$

You can confirm this by using a tool like Desmos. See below.

------------------------------------------------------------------------

Problem 8

The endpoints of the diameter are (-2,1) and (8,25)

The center is the midpoint of these endpoints.

The midpoint of the x coordinates is (-2+8)/2 = 3

The midpoint of the y coordinates is (1+25)/2 = 13

The center is (h,k) = (3,13)

Now find the distance from the center to one of the points on the circle, let's say to (8,25)

$(x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\$