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3 years ago

Does ABC appear to be the same size and shape as

Mathematics

1 answer:

nikitadnepr [17]3 years ago

6 0

Answer:

most probably number D

Step-by-step explanation:

Z and X is angled differently compared to A and C making it different sizes.

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Brilliant_brown [7]

Answer:

$f(x)=\sqrt{x-1}$ -----> Graph C

$f(x)=\sqrt{x}$ -----> Graph A

$f(x)=\sqrt{x}-1$ -----> Graph B

$f(x)=-\sqrt{x}$ -----> Graph D

$f(x)=-\sqrt{x-1}$ -----> Graph E

Step-by-step explanation:

we have

case a) $f(x)=\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case b) $f(x)=\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case c) $f(x)=\sqrt{x}-1$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [-1,∞)

All real numbers greater than or equal to -1

case d) $f(x)=-\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

case e) $f(x)=-\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

5 0

3 years ago

Elimination method<br> 4x-2y=48<br> 2x+6y=-18

yulyashka [42]

$4x-2y=48\\ \\ 2x+6y=-18$

First we'll multiply the second equation by -2

$-2\cdot (2x+6y)=-2\cdot -18\\ \\ -4x-12y=36$

Now let's add the new equation and the first one.

$(4x-2y)+(-4x-12y)=48+36\\ \\ 4x-4x-2y-12y=48+36\\ \\ -14y=84\\ \\ y=\frac { 84 }{ -14 } \\ \\ y=-6$

We found y's value. We'll plug it in one of the equations to find x's value.

$y=-6\\ \\ 4x-2y=48\\ \\ 4x-2\cdot -6=48\\ \\ 4x+12=48\\ \\ 4x=48-12\\ \\ 4x=36\\ \\ x=\frac { 36 }{ 4 } \\ \\ x=9$

Solution ;

(9, -6)

4 0

3 years ago

Read 2 more answers

4/25+1/5 in simplest form

Evgen [1.6K]

It would be 9/25. First you find the common denominator which is 25 so then you convert 1/5 to 5/25 so now you have 4/25+5/25 which you would get 9/25 and since 9 can only be divided be 3 and 25 can't your answer would be 9/25.

3 0

4 years ago

Please help - ACT math practice test

bezimeni [28]

Answer:

Step-by-step explanation:

Let's look at the equation.

$h=-at^{2}+bt+c$