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4 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

Savatey [412]4 years ago

4 0

A' = (-2, 1)
B' = (1, 0)
C' = (-1, 0)

natta225 [31]4 years ago

4 0

A'= -2, -1
B'= 1, 0
C'=-1,0

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Usain Bolt ran the 2012 Olympic 100-meter race in 9.63 seconds. If he runs at this rate on a road with a speed limit of 25 miles

djverab [1.8K]

Answer:

3874.5342

Step-by-step explanation:

9.63 divide by 100 = 0.0963

0.0963x40 234= 3874.5342

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3 years ago

Read 2 more answers

Plz, help ASAP!!!!!!<br> WILL MARK BRAINLIEST IF YOUR ANSWER IS CORRECT!!!!!!

Mekhanik [1.2K]

Blank 1: Given

Blank 2 : AD is parallel to BC

Blank 3: <3≅<2

Blank 4: Transitive property of convergence

Blank 5: Definition of Bisect

Step-by-step explanation:

We need to fill in the blanks from the given options.

Blank 1:

ABCD is a parallelogram

This is given

So, blank 1: Given

Blank 2:

A parallelogram have parallel sides i.e in the given parallelogram AD is parallel to BC and AB is parallel to DC

So, Blank 2 : AD is parallel to BC

Blank 3:

We are given <3≅<2

So, Blank 3: <3≅<2

Blank 4:

We know, <1≅<3 and <3≅<2 using transitive property (A≅B and B≅C then A≅C)

So, <1 ≅ <2 is due to transitive property of convergence

Blank 4: Transitive property of convergence

Blank 5:

DC bisects <ADE

A bisect is an line or angle that divides into exactly two parts.

So, Blank 5: Definition of Bisect

Keywords: parallelogram, bisection

Learn more about parallelogram at:

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4 0

3 years ago

Draw a model to help solve 5/6 + 1/4. Write your anserw as a mixed number

deff fn [24]

Answer:

Step-by-step explanation:

7 0

2 years ago

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

HEEELP

Drupady [299]

Answer: (c)

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}$

Here, $\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)$

To get $f\left(g(x)\right)$ , replace $x$ in $f(x)$ by $g(x)\ \text{i.e. by}\ \sqrt{x+5}$

$\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)$

Using $(i)$ and $(ii)$ it can be concluded that the domain of $f\left(g(x)\right)$ is all real numbers except 0.

Therefore, its domain is given by

$x\in [-5,4)\cup (4,\infty)$

Option (c) is correct.

5 0

3 years ago