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4 years ago

Find the range and the domain of the function below

Mathematics

1 answer:

Tasya [4]4 years ago

7 0

D :30
R :90
That's the answer hehe

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Which figure shows a reflection of pre-image QRS over the line t?

Igoryamba

Answer:

A, because its reflecting off of the y axis

Step-by-step explanation:

I have proof links, and I took the test

4 0

3 years ago

Maya paid $11.88 for a 2.38-pound bag of shrimp at one store. The following week, she paid $26.47 for a 5.08-pound bag at anothe

lbvjy [14]

Answer:

2.38 lb bag = 4.99/lb 5.08 lb bag = 5.21/lb The 2.38 is the better buy

Step-by-step explanation:

11.88/2.38 = 4.99

26.47/5.08 = 5.21

4.99 is cheaper than 5.21 so the 2.38 lb bag is better buy

6 0

3 years ago

I need help on .5 and .6

Dovator [93]

5. 44 (forty four)
6. 58

8 0

3 years ago

A government agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted

mylen [45]

let n be the number of times letter be found (1800), p be the success rate (0.068), and q be the failure rate (1 - 0.068 = 0.932)

mean = np = 1800 x 0.068 = 122.4

standard deviation = √(npq) = √(1800 x 0.068 x 0.932) = 10.6807414

4 0

3 years ago

I have nooooo clue, please help

Aleksandr-060686 [28]

Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

$y=(1/3)x^{2}$

$a=(1/3)$

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right

3 0

3 years ago