All categories

3 years ago

Line JK bisects line LM at point J. Find JM if LJ=23 centimeters

Mathematics

2 answers:

sveta [45]3 years ago

5 0

Answer:

The length of JM is 23 centimeters.

Step-by-step explanation:

It is given that Line JK bisects line LM at point J. It means J is the midpoint of line LM.

A midpoint divides a line in two equal parts.

J is midpoint of LM, so we can say that the distance between L ans J, J and M are same.

$LJ=JM$

It is given that LJ=23 centimeters. Substitute LJ=23 in the above equation to find the value of JM.

$23=JM$

Interchanging the sides, we get

$JM=23$

Therefore the length of JM is 23 centimeters.

vlada-n [284]3 years ago

4 0

Ling JM is also 23 centimeters because line LJ and JM are a line but broken into two segments

You might be interested in

Simplify (x - y + 1) - (x + y - 1)

andrey2020 [161]

First, distribute the minus sign across the second set. So you have:

x - y + 1 - x - y + 1

Now, we combine like terms:
x-x=0
-y-y=-2y
1+1=2

So we have -2y+2

Hope that helps.

5 0

4 years ago

Which table represents an arithmetic sequence?

andriy [413]

An arithmetic sequence is a sequence of integers with its adjacent terms differing with one common difference. The table that represents an arithmetic sequence is the third table.

<h3>What is arithmetic sequence?</h3>

An arithmetic sequence is a sequence of integers with its adjacent terms differing with one common difference.

The explicit formula for any arithmetic series is given by the formula,

aₓ = a₁ + (x-1)d

where d is the difference and a₁ is the first term of the sequence.

For the table to be in an arithmetic sequence, the difference between any two consecutive terms must be equal.

For the first table, the difference between the first two terms is -6, while for the next two terms it is -12. Thus, it is not an arithmetic sequence.
In the second table, the difference between the first two terms is 2 while the difference between the next two terms is 4. Thus, it is not an arithmetic sequence.
In the third table, the difference between the first two terms is 1.4, the difference between the next two terms is 1.4. Also, it last two terms the difference is 1.4. Thus, it is an arithmetic sequence.

Hence, the table that represents an arithmetic sequence is the third table.

Learn more about Arithmetic Sequence:

brainly.com/question/3702506

#SPJ1

4 0

2 years ago

y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the

vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

$y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},$

where $c_1, c_2 \in \mathbb{R}$ . We now want to find a solution $y$ such that $y(-1)=3$ and $y'(-1)=-3$ . Therefore, all we need to do is find the constants $c_1$ and $c_2$ that satisfy the initial conditions. For the first condition, we have: $y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.$

For the second condition, we need to find the derivative $y'$ first. In this case, we have:

$y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.$

Therefore:

$y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.$

This means that we must solve the following system of equations:

$\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.$

If we add the equations above, we get:

$\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.$

If we now substitute $c_1 = 0$ into either of the equations in the system, we get:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

$\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.$

Indeed, we can see that:

$y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3$

$y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,$

which do correspond to the desired initial conditions.

3 0

3 years ago

What is the answer to this question? <br> (The one in the middle)

Delicious77 [7]

Let f=number of field goals and a=number of points after.
a+f=38
a+3f=70
.
a+3f=70

a+ f=38
-------subtract
2f=32
.
f=16 field goals he kicked last season.

4 0

4 years ago

What is the simplified form of the following expression (5/4)^3

Vlad [161]

★ EXPONENTIALLY RESOLVED ★

(5/4)^3

= 5 ( 5 ) ( 5 ) / 4 ( 4 ) (4 )

= 25 ( 5 ) / 16 ( 4 )

= 125 / 64

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★

3 0

3 years ago