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3 years ago

Pls help I don’t get it

Mathematics

2 answers:

nirvana33 [79]3 years ago

8 0

Well you are trying to find x, so you would divide 5/8 by itself and on the other side.

(5/8x / 5/8) = (10 / 5/8)

x=16

Hope this helps!

Firdavs [7]3 years ago

5 0

Here you gooooo!!!!!!!!

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Which graph shows dilation?

Evgesh-ka [11]

Answer:

The First Image

Step-by-step explanation:

A dilation is a type of transformation that enlarges or reduces a figure, therefore in the first Image we see an increase in scale factor of 2.

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3 years ago

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For A (1, –1), B (–1, 3), and C (4, –1), find a possible location of a fourth point, D, so that a parallelogram is formed using

katrin [286]

The possible coordinates for the vertex D of the parallelogram could be (-4, 3)

It is given that A(1, –1), B (–1, 3), and C (4, –1) are the three vertices of the parallelogram.

Let us assume that the vertices are in the order A, C, B, and D where the coordinates of D are (a, b).

In this scenario, if we join AB and CD, they will become the diagonals of the parallelogram ABCD.

According to the properties of a parallelogram, diagonals bisect each other.

Hence, mid-point of AB = mid-point of CD

Now, according to the mid-point theorem, if mid-point of AB is given as (x,y), then,

x = (x₁ + x₂)/2 and y = (y₁ + y₂)/2

Here, for AC,

x₁ = 1, y₁ = -1

x₂ = -1, y₂ = 3

Then, x = ( 1 - 1)/2 and y = (-1 + 3)/2

(x, y) ≡ (0, 1) ............. (1)

Since (x, y) is also the mid-point of CD, we also have,

x₁ = 4, y₁ = -1

x₂ = a, y₂ = b

Then, x = (4 + a)/2 and y = (-1 + b)/2

(x, y) ≡ ((4 + a)/2, (-1 + b)/2) ................... (2)

From (1) and (2),

(4+a)/2 = 0 and (-1+b)/2 = 1

4+a = 0 and (-1+b) = 2

a = -4 and b = 3

Hence, the fourth vertex D of the parallelogram can be possibly located at (-4, 3)

Learn more about a parallelogram here:

brainly.com/question/1563728

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8 0

2 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

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3 years ago

Simplify. 5 - {-(-2)} A)-3 B) 3 C)-7 D) 7

Fed [463]

5 - {-(-2)}
5 - {+2} (negative + negative = positive)
5 - 2 (negative + positive = negative)
3

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3 years ago

In the drawing below, label the distances given for the peak you chose

earnstyle [38]

Notice the larger triangle is a right-triangle, and the small one is also a right-triangle. Also notice the "angle tickmark" on each. So both angles have a pair of angles that are equal on each.

the triangles are similar by AA.

3 0

3 years ago

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