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3 years ago

What is the range of the equation ?

Mathematics

1 answer:

nata0808 [166]3 years ago

4 0

Option D:

all real numbers

Solution:

Definition of domain:

The domain of the function is the set of all possible

x-values (independent variable).

Definition of range:

The range of a function is the set of all possible resulting values

of y-values (dependent variable).

i.e The range of a function is the resulting y-values we get after substituting all the possible x-values.

Given function: $y=\log_8x$

Domain of y: (0, ∞) (or) {x | x > 0}

Range of y: (–∞, ∞) (or) {y | y > R}

i.e all real numbers

Option D is the correct answer.

Hence the range of y is all real numbers.

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The planning committee at school has 10 members. Exactly four of these members are teachers. A four-person subcommittee with at

Kipish [7]

Answer:

865

Step-by-step explanation: The set of possibilities are:

Subcommittee is of 4 people according to:

4 teachers 0 other member total possibilities 1

3 teachers 1 teachers C₄,₃ = 4 and other member 6 6*4 = 24

2 teachers 2 C₄,₂ = 4!/2! = 12 2 (om) and C₆,₂ = 6!/4! = 30

30*12 = 360

1 teacher 3 (om) 4 and C₆,₃ = 6!/3! = 120

120*4 = 480

Total ways of committee 1 + 24 + 360 + 480 = 865

4 0

3 years ago

To the nearest tenth, which is the perimeter of ABC. Geometry

qwelly [4]

Answer:

23.6

Step-by-step explanation:

Finding AC:

Cos 61 = $\frac{adjacent}{hypotenuse}$

0.48 × 10 = Adjacent

AC = 4.8

Now, CB:

Cos 29 = $\frac{adjacent}{hypotenuse}$

0.87 × 10 = CB

CB = 8.8

The perimeter:

=> 10+4.8+8.8

=> 23.6

3 0

3 years ago

Read 2 more answers

Which equation represents the circle described? (in picture)

Sedaia [141]

You are right. In circle equation we have =r² so the correct answer will be C od D. But we have to calculate the center of a circle.

$x^2+y^2-8x-6y+24=0\quad|-24\\\\(x^2-8x)+(y^2-6y)=-24$

Complete the square:

$x^2-8x+a^2\implies a=\dfrac{8x}{2x}=4\implies x^2-8x+4^2\\\\\\y^2-6y+b^2\implies b=\dfrac{6y}{2y}=3\implies y^2-6y+3^2$

And we have:

$(x^2-8x)+(y^2-6y)=-24\\\\(x^2-8x+4^2)-4^2+(y^2-6y+3^2)-3^2=-24\\\\
(x-4)^2+(y-3)^2-16-9=-24\\\\(x-4)^2+(y-3)^2-25=-24\quad|+25\\\\\boxed{(x-4)^2+(y-3)^2=1}$

So the answer is C (the same left side of equation).

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Answer:

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kykrilka [37]

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