Pwease halp uwu im so dumb...

Which of the following is the graph of y=- square root x+1

maks197457 [2]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$y=-\sqrt{x+1}$

Find the intercepts

The y-intercept is the value of y when the value of x is equal to zero

so

For x=0 -----> $y=-\sqrt{0+1}=-1$

The y-intercept is the point (0,-1)

The x-intercept is the value of x when the value of y is equal to zero

so

For y=0 -----> $0=-\sqrt{x+1}$ -----> $x=-1$

The x-intercept is the point (-1,0)

Find the domain

we know that

The radicand must be positive

so

$x+1\geq0$

$x\geq-1$

All real numbers greater than or equal to -1

The range is all real numbers less than or equal to -1

using a graphing tool

see the attached figure

4 0

3 years ago

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1/2 x + 13 = 9 (x Is Variable)

Elza [17]

Answer:

x= -8

Step-by-step explanation:

1/2x + 13=9

- move 13 to the other side

1/2x= 9-13

-simply 9-13

1/2x= -4

-multiply 2 on both sides

1x=-8

x=-8

3 0

3 years ago

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Simplify the expression below. 12b + 86 + 5b 0 25 + b O 25(3b) 25b 3b

Alja [10]

(1): "^-5" was replaced by "^(-5)". 3 more similar replacement(s)

STEP

1

:

Equation at the end of step 1

((25•(b-1))•(c-3))

(((12•(b4))•(c-6))•——————————————————)•((2•5b(-5))•c)

((15•(b3))•(c4))

STEP

2

:

Equation at the end of step

2

:

((25•(b-1))•(c-3))

(((12•(b4))•(c-6))•——————————————————)•(2•5b(-5)c)

((3•5b3)•c4)

STEP

3

:

Equation at the end of step

3

:

(52b(-1)•c(-3))

(((12•(b4))•(c-6))•———————————————)•(2•5b(-5)c)

(3•5b3c4)

STEP

4

:

52b(-1)c(-3)

Simplify ————————————

(3•5b3c4)

Dividing exponential expressions :

4.1 b(-1) divided by b3 = b((-1) - 3) = b(-4) = 1/b4

Dividing exponential expressions :

4.2 c(-3) divided by c4 = c((-3) - 4) = c(-7) = 1/c7

Dividing exponents:

4.3 52 divided by 51 = 5(2 - 1) = 51 = 5

Equation at the end of step

4

:

5

(((12•(b4))•(c-6))•—————)•(2•5b(-5)c)

3b4c7

STEP

5

:

Equation at the end of step

5

:

5

(((22•3b4) • c(-6)) • —————) • (2•5b(-5)c)

3b4c7

STEP

6

:

Canceling Out :

6.1 Canceling out b4 as it appears on both sides of the fraction line

Dividing exponential expressions :

6.2 c(-6) divided by c7 = c((-6) - 7) = c(-13) = 1/c13

Canceling Out:

6.3 Canceling out 3 as it appears on both sides of the fraction line

Equation at the end of step

6

:

20

——— • (2•5b(-5)c)

c13

STEP

7

:

Multiplying exponents:

7.1 22 multiplied by 21 = 2(2 + 1) = 23

Multiplying exponents:

7.2 51 multiplied by 51 = 5(1 + 1) = 52

Dividing exponential expressions :

7.3 c1 divided by c13 = c(1 - 13) = c(-12) = 1/c12

Final result :

200

—————

b5c12 Answer:

Step-by-step explanation:

5 0

3 years ago

An identification code is to consist of 3 letters followed by 2 digits. Determine the following

kirill [66]

Well let's see:

The first letter can be any one of 26 .
   For each one . . .
The second letter can be any one of the remaining 25.
   For each one . . .
The third letter can be any one of the remaining 24.
   For each one . . .
The two digits can be any number from 01 to 98 ...
   except 11, 22, 33, 44, 55, 66, 77, or 88. (No repetition.)
There are 90 of them.

So the total number of possibilities is (26 · 25 · 24 · 90) .

When I multiply that out, I get 1,404,000 .

I don't know how you got your number, so I can't comment on your
method, but I did find something interesting about your number:

If I assume that you did the three letters the same way I did, then
if I divide your number by (26·25·24), the quotient will show me
how you handled the two digits.

   1,263,600 / (26·25·24) = 81 .

That's very intriguing, because it's so close to the 90 sets of digits
that I used. But I don't know what it means, or if it means anything
at all.

7 0

3 years ago

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$