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Alex17521 [72]
3 years ago
8

What is the distance around a triangle with the sides measuring 2⅛, 3½ and 2½?

Mathematics
1 answer:
Marta_Voda [28]3 years ago
7 0
So when you have data given this kind of form... you should convert to decimal first.
So that would be:
a ≈ 2.12
b = 3.5
c = 2.5
Now just place the simple perimeter formula, and sum all sides.
You might be interested in
19. which one of the following is not equal to the rest? a 2% of 150 b. % of 400 b 3 c. 6% of 60 d. 69 of 50​
Irina18 [472]

Answer:

i think it's c

Step-by-step explanation:

a. <u>2</u><u>/</u><u>1</u><u>00</u><u> </u><u>×</u><u> </u><u>1</u><u>5</u><u>0</u><u>=</u><u> </u><u>3</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>b</u><u>. </u><u> </u><u>100×</u><u>4</u><u>0</u><u>0</u><u>=</u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u>c</u><u>. </u><u> </u><u> </u><u>6</u><u>/</u><u>100 </u><u>×</u><u>6</u><u>0</u><u>=</u><u> </u><u>1</u><u>.</u><u>2</u><u> </u><u>×</u><u>3</u><u>=</u><u>3</u><u>.</u><u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>d</u><u>. </u><u>6</u><u>9</u><u>×</u><u>5</u><u>0</u><u>=</u><u>3</u><u>4</u><u>0</u><u>0</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>so </u><u>answer</u><u> </u><u>c</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u>

5 0
3 years ago
Can someone help me do part two please? It’s very important send a picture or something. I don’t even care if you tell me the st
Nataly_w [17]
<h3>Explanation:</h3>

1. "Create your own circle on a complex plane."

The equation of a circle in the complex plane can be written a number of ways. For center c (a complex number) and radius r (a positive real number), one formula is ...

  |z-c| = r

If we let c = 2+i and r = 5, the equation becomes ...

  |z -(2+i)| = 5

For z = x + yi and |z| = √(x² +y²), this equation is equivalent to the Cartesian coordinate equation ...

  (x -2)² +(y -1)² = 5²

__

2. "Choose two end points of a diameter to prove the diameter and radius of the circle."

We don't know what "prove the diameter and radius" means. We can show that the chosen end points z₁ and z₂ are 10 units apart, and their midpoint is the center of the circle c.

For the end points of a diameter, we choose ...

  • z₁ = 5 +5i
  • z₂ = -1 -3i

The distance between these is ...

  |z₂ -z₁| = |(-1-5) +(-3-5)i| = |-6 -8i|

  = √((-6)² +(-8)²) = √100

  |z₂ -z₁| = 10 . . . . . . the diameter of a circle of radius 5

The midpoint of these two point should be the center of the circle.

  (z₁ +z₂)/2 = ((5 -1) +(5 -3)i)/2 = (4 +2i)/2 = 2 +i

  (z₁ +z₂)/2 = c . . . . . the center of the circle is the midpoint of the diameter

__₁₂₃₄

3. "Show how to determine the center of the circle."

As with any circle, the center is the <em>midpoint of any diameter</em> (demonstrated in question 2). It is also the point of intersection of the perpendicular bisectors of any chords, and it is equidistant from any points on the circle.

Any of these relations can be used to find the circle center, depending on the information you start with.

As an example. we can choose another point we know to be on the circle:

  z₄ = 6-2i

Using this point and the z₁ and z₂ above, we can write three equations in the "unknown" circle center (a +bi):

  • |z₁ - (a+bi)| = r
  • |z₂ - (a+bi)| = r
  • |z₄ - (a+bi)| = r

Using the formula for the square of the magnitude of a complex number, this becomes ...

  (5-a)² +(5-b)² = r² = 25 -10a +a² +25 -10b +b²

  (-1-a)² +(-3-b)² = r² = 1 +2a +a² +9 +6b +b²

  (6-a)² +(-2-b)² = r² = 36 -12a +a² +4 +4b +b²

Subtracting the first two equations from the third gives two linear equations in a and b:

  11 -2a -21 +14b = 0

  35 -14a -5 -2b = 0

Rearranging these to standard form, we get

  a -7b = -5

  7a +b = 15

Solving these by your favorite method gives ...

  a +bi = 2 +i = c . . . . the center of the circle

__

4. "Choose two points, one on the circle and the other not on the circle. Show, mathematically, how to determine whether or not the point is on the circle."

The points we choose are ...

  • z₃ = 3 -2i
  • z₄ = 6 -2i

We can show whether or not these are on the circle by seeing if they satisfy the equation of the circle.

  |z -c| = 5

For z₃: |(3 -2i) -(2 +i)| = √((3-2)² +(-2-i)²) = √(1+9) = √10 ≠ 5 . . . NOT on circle

For z₄: |(6 -2i) -(2 +i)| = √((6 -2)² +(2 -i)²) = √(16 +9) = √25 = 5 . . . IS on circle

4 0
3 years ago
A carpenter has a wooden dowel that is 72 centimeters long. She wants to cut it into 2 pieces so that one piece is 5 times as lo
Jobisdone [24]
Let's say that the shorter piece is x and the longer piece is 5x (because it's 5 times as long). x+5x=72 and 6x=72. Dividing both sides by 6, we get that x=12, and 72-12=the length of the other piece (since there are only 2 pieces)=60
6 0
3 years ago
Given that α and β are the roots of the quadratic equation <img src="https://tex.z-dn.net/?f=2x%5E%7B2%7D%20%2B6x-7%3Dp" id="Tex
siniylev [52]

Answer:

\large \boxed{\sf \ \ \ p=-11 \ \ \ }

Step-by-step explanation:

Hello,

\alpha \text{ and } \beta \text{ are the roots of the following equation}

   2x^2+6x-7=p

It means that

   2\alpha^2+6\alpha-7=p \\\\2\beta ^2+6\beta -7=p \\\\

And we know that

\alpha= 2\cdot \beta

So we got two equations

   2(2\beta)^2+6\cdot 2 \cdot \beta -7=p \\\\8\beta^2+12\beta -7=p\\\\ and \ 2\beta ^2+6\beta -7=p \ So \\\\\\8\beta^2+12\beta -7 = 2\beta ^2+6\beta -7\\\\6\beta^2+6\beta =0\\\\\beta(\beta+1)=0\\\\ \beta =0 \ or \ \beta=-1

For \beta =0, \ \ \alpha =0, \ \ p = -7

For \beta =-1, \ \ \alpha =-2, \ \ p= 2-6-7=-11, \ p=2*4-12-7=-11

I assume that we are after two different roots so the solution for p is p=-11

b) \alpha +2 =-2+2=0 \ and \ \beta+2=-1+2=1

So a quadratic equation with the expected roots  is

x(x-1)=x^2-x

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

3 0
3 years ago
Find a and b so that f(x) = x^3 + ax^2 + b will have a critical point at (2,3).
Digiron [165]

Using the critical point concept, it is found that a = -3 and b = 7.

<h3>What are the critical points of a function?</h3>
  • The critical points of a function are the values of x for which:

f^{\prime}(x) = 0

In this problem, the function is:

f(x) = x^3 + ax^2 + b

Hence, the derivative is:

f^{\prime}(x) = 3x^2 + 2ax

Then:

f^{\prime}(x) = 0

3x^2 + 2ax = 0

x(3x + 2a) = 0

x = 0

3x + 2a = 0

3x = -2a

x = -\frac{2a}{3}

Since the critical point is at x = 2, we have that:

-\frac{2a}{3} = 2

-2a = 6

2a = -6

a = -3

Then:

f(x) = x^3 - 3x^2 + b

Critical point at (2,3) means that when x = 2, y = 3, then:

3 = 2^3 - 3(2)^2 + b

8 - 12 + b = 3

b = 7

You can learn more about the critical point concept at brainly.com/question/2256078

4 0
2 years ago
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