Help on Question 14 , 15

Write an addition equation that can help you find 9-3

Alborosie

X+3=9
-3 -3
________
X=3

7 0

3 years ago

Factor 4a2 - 25. Type your answer in the space provided. Type your answer in the space provided. Do not type spaces in your answ

Dvinal [7]

Answer: The answer to this question is (2a-5)(2a+5)

Step-by-step explanation:

Because first, you have to write the quadratic equation 4a^2-25 in exponential form

Then, you calculate the product by multiplying the terms with equal exponents by multiplying their bases

(2a)^2-5^2

And then last but not least, using a^2-b^2=(a-b)(a+b), factor the expression

(2a-5)(2a+5) and then there's your answer and your explanation!!!!!

8 0

3 years ago

This is 4 questions.

jeka94

1) a. 15% decrease
2) a. $79.20
3) b. 14.3% increase
4) a. $4.68

#1 explanation:
1435 - 1220 = 215
215/1435 = 0.1498
0.1498 x 100 = 14.98 = 15%

#2 explanation:
220 x 0.6 = 132
220-132=88
88 x 0.1 = 8.8
88-8.8 = 79.2 $

#3 explanation:
32-28=4
4/28=0.1428
0.1428 x 100 = 14.28 = 14.3%

#4 explanation:
275 x 0.15 = 41.25
275-41.25 = 233.75
233.75 x 0.02 = 4.675 = 4.68 $

4 0

3 years ago

Read 2 more answers

One of the students is chosn at random. What is the probability of picking a student who sciored above 65

Andrews [41]

Answer: 7 / 18

Step-by-step explanation:

There were 18 students.

Out of them, the scores above 65 are:

73, 73, 69, 68, 71, 76, 70

There are 7 of them.

Probability of picking someone who scored higher than 65 is:

= Number of people who scored above 65/ Total number who took the test

= 7 / 18

6 0

3 years ago

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

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