All categories

2 years ago

Please help me on this!!! I would appreciate it so much!!!

Mathematics

1 answer:

kap26 [50]2 years ago

4 0

Y=4.7x+2.1
v=1.6x-5.4

=
(x, y) = ( -75/31, - 1437/155)

You might be interested in

A car dealer recently sold five cars for the following profits: $10,126, $9,999, $12,398, $12,007, and $4,567. What was the mean

Vikentia [17]

Mean or Average is all of the sums divided by how many of the choices are there, so all of 10,126+9,99+12,398+12,007+4567=49097 and divided by all the choices you have which is 5 is 9819.4

7 0

3 years ago

Read 2 more answers

What is the value of the expression?<br> 2 1/4 divided by 1 2/5

ad-work [718]

Answer:

45/28

Step-by-step explanation:

Convert them to mixed fractions

(2 * 4 + 1) / 4 ÷ (1 * 5 + 2) / 5

8 + 1 / 4 ÷ 5 + 2 / 5

9/4 ÷ 7/5

Flip the second fraction and flip the sign - to multiply.

9/4 × 5/7

9 × 5 = 45

4 × 7 = 28

45/28

This could be left as a fraction or converted to a number:

1.60714285714

4 0

3 years ago

Read 2 more answers

Question 7 of 10

VLD [36.1K]

Answer:

It is increasing

The y-intercept is (0, 3).

The domain of F(x) is all real numbers.

7 0

2 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

3 years ago

Read 2 more answers

What’s 1 plus 3 and write a paragraph on why that’s your awnser. Include details

aivan3 [116]

Answer:

1+3=4

Step-by-step explanation:

why the hll do u need a paragraph on that ur in highschool, just why

3 0

3 years ago

Read 2 more answers