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2 years ago

What is the value of y? 1 2 7 8

Mathematics

1 answer:

SVETLANKA909090 [29]2 years ago

4 0

Answer:

Step-by-step explanation:

$8y+18=12y-10\\ 12y-8y=10+18\\ 4y=28\\ y=\frac{28}{4} \\ y=7$

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X - 10+ 9x = - 1 + 10x

Karo-lina-s [1.5K]

Answer:

<h2></h2><h2><u><em>the equation is false</em></u></h2>

Step-by-step explanation:

x - 10+ 9x = - 1 + 10x

-10 + 1 = -x - 9x + 10x

-9 = 0

the equation is false

7 0

9 months ago

Which graph best represents a quadratic function that has only one zero?

GenaCL600 [577]

Answer:

Step-by-step explanation:

A has 2 it crosses the x axis 2x

B Has 1.

C has 2

D has none since it never touched the x axis

8 0

1 year ago

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Find the area of a regular triangle with side length of 15.5 inches

Ganezh [65]

Answer:

Brainliest pls

Step-by-step explanation:

A≈104.03in²

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1 year ago

One tile yellow, 2 tiles blue, and 3 tiles are purple. What fraction of tiles are yellow or purple

Harlamova29_29 [7]

3 plus 1 equals 4 so its 4/6. When you simplify it the answer is 2/3

3 0

2 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

2 years ago