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3 years ago

Find the domain and range of y= -|x|-1

Mathematics

2 answers:

Debora [2.8K]3 years ago

7 0

R: (-inf, 0]
D: all real #'s

Mnenie [13.5K]3 years ago

4 0

The Domain is all real numbers, and since the graph is translated 1 down and flipped across the x-axis, the Range is all real numbers less than or equal to -1.

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Giving brainleist help pls

Tomtit [17]

Answer:B

Step-by-step explanation:100% correct

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2 years ago

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Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

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2 years ago

PLEASE ANSWER ASAP FOR BRAINLEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Trava [24]

Answer: Should be 150mm

Step-by-step explanation:

Find the surface area of each figure. Round answers to the nearest tenth, if necessary

2 mm-

2/5 mm

10 mm

3 mm

75 mm2

150 mm2

92 mm2

86 mm2

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3 years ago

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Adams says that the expression 52-3y is equal to 20 when y = 2 explain adams answer is incorrext

N76 [4]

Adam is incorrect because if you substitute the value of y (which is 2) on the expression 52-3y=20, the statement is false since 46 is obviously not equal to 20.

52-3(2)=20
52-6=20
46 = 20

FALSE

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3 years ago

Ms. Francis is planning a birthday party for her daughter. There will be 22 children at the party, and in order to seat them all

Oxana [17]

Ms Francis would need to rent 6 tables in order to get everyone their own side.

She could have all the children, sitting at their tables, having their own side, but the 6th table just having the two other kids.
Ms Francis shouldn’t be having such a large party in the midst of a pandemic.

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3 years ago