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topjm [15]
3 years ago
13

Is the figure defined by points H, I, J, and K a rhombus?

Mathematics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

b

Step-by-step explanation:

You might be interested in
Determine the missing side length:
ludmilkaskok [199]
The answer is:  "3" .
_______________________________
Use the Pythagorean theorem (for right triangles):

a² + b² = c² ;

in which "a = "side length 1" (unknown; for which we which to solve);

               "b" = "side length 2" = "√3" (given in the figure) ;

               "c" = "length of hypotenuse" = "2√3" (given in the figure);
_____________________________________________________
a² + b² = c² ; 

a² = c² − b² ;

Plug in the known values for "c" and "b" ;

a² = (2√3)²  −  (√3)² ;

Simplify:

(2√3)² = 2² * (√3)² = 2 * 2 * (√3√3)  = 4 * 3 =  12 .

 (√3)²  = (√3√3) = 3 .

a²  = 12 − 3 = 9 .

a² = 9

Take the "positive square root" of EACH SIDE of the equation; to isolate "a" on one side of the equation;  & to solve for "a" ; 

+√(a²)  = +√9 ;

     a = 3 . 
____________________________________
The answer is:  "3" .
____________________________________
3 0
3 years ago
1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

3 0
3 years ago
Suppose that a college student is taking three courses: a two-credit course, a three-credit course, and a four-credit course. Th
forsale [732]

Answer:

The overall expected grade point average for the semester is 3.16/3.1 or 3.2 if you round it up.

Step-by-step explanation:

To get the average, you need to add up all the numbers then divide the sum by how many numbers you added.

3.5 + 3.0 + 3.0 = 9.5

9.5 / 3 = 3.16

4 0
3 years ago
Read 2 more answers
Kaley cuts half of a loaf of bread into 4 equal parts. What fraction of the whole loaf does each of the 4 parts represent?
morpeh [17]

Answer:

The answer is 1/4

Step-by-step explanation:

When you split something into fourths, you each get 1/4 of it.

6 0
3 years ago
Read 2 more answers
If k is parallel to l find the value of a and the value of b.
kipiarov [429]

Answer:

a = 30

b = 40

C

Step-by-step explanation:

This requires that you use a proportion.

a / (a + 15) = 40 / 60

The small triangle's sides are in proportion to the large triangles sides.

Reduce the right. Divide top and bottom by 20

a/(a + 15) = 40/20 // 60/20

a/(a + 15) = 2/3                    

Cross multiply

3a = 2(a + 15)            Remove the brackets on the right.

3a = 2a + 30

Subtract 2a from both sides

3a-2a = 30

a = 30

Find B

b is done exactly the same way.

b/(b+20) = 2/3

3b = 2b + 40

3b - 2b = 40

b = 40

3 0
4 years ago
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