What Two decimals are equivalent to 0.9

How to find the value of x and z

Setler79 [48]

Answer:

z = 86 °

and, x = 2

Step-by-step explanation:

We know that a straight line is a 180 degree-angle. So, we know that

94 + z must equal 180, because they are the only two angles that make up the straight line.

from this information, (that 94 + z = 180) we can easily find z.

(this is typically an intuitive method of subtracting 94 from 180)

94 + z = 180

-94 -94

z = 86 °

-- we know that the relation between z and (10x + 74) must also be equal to 180, as they make up a straight line

if z = 86, then we can know that (10x + 74) must be equal to

(10x + 74) + 86 = 180

- 86 -86

(10x + 74) = 94

10x + 74 = 94

- 74 - 74

10 x = 20

÷ 10 ÷ 10

x = 2

so, z = 86 °

and, x = 2

(I am assuming that this "unprotected placement assessment" has already taken place, and you would like to understand a question. Or, this is a practice assessment. Because there is no point to cheating on a placement assessment--you will just skip over material you don't know. And, it wouldn't align with Brainly's honor policy. Hope you learn from this!!)

7 0

2 years ago

Ramona had 8 3/8 in. of ribbon. She used 2 1/2 in. for an art project.How many inches of ribbon does she have left?

disa [49]

$8\frac{3}{8}= \frac{67}{8}$

$2\frac{1}{2} = \frac{5}{2} = \frac{20}{8}$

$\frac{67}{8}- \frac{20}{8} = \frac{47}{8} =5 \frac{7}{8} in$

8 0

4 years ago

Read 2 more answers

Solve. 7x − 2 = −23 x =

Zielflug [23.3K]

7x-2=-23 add 2 to both sides to get 7x by itself
7x=-23+2 now add the right side
7x=-21 now divide both sides by 7
x=-21/7
x=-3

7 0

4 years ago

Read 2 more answers

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$