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Stolb23 [73]
2 years ago
9

Which statements accurately describe the function f(x) = 3(16)^3/4? Check all that apply.

Mathematics
2 answers:
riadik2000 [5.3K]2 years ago
8 0

Answer:

a,c,e

Step-by-step explanation:

uysha [10]2 years ago
6 0

Answers:

These are the statements that apply:

The initial value is 3.

The range is y >0.

The simplified base is 8.

Explanation:

1) Given expression:

f(x)=3(16)^{\frac{3}{4} x

2) Check every statement:

a) The initial value is 3?

initial value ⇒ x = 0 ⇒

f(0)=3(16)^{0}=3(1)=3

∴ The statement is right.

b) The initial value is 48?

Not, as it was already proved that it is 3.

c) The domain is x > 0?

No, because the domain of the exponential functions is all the Real numbers.

d) The range is y > 0?

That is correct, the exponential function is continuous, and monotonon increasing.

The limit when x → - ∞ is zero, but y never reaches zero, and the limit when x → ∞ is + ∞, meaning that the range is y > 0.

e) The simplified base is 12?

This is how you simplify the base:

3(16)^{\frac{3}{4} x}=3{{(16}^{(3/4)})}^x=3(16^{3/4}})^{x}=3((2^4)^{3/4})^x=3(2^3)^x=3(8)^x

Which shows that the simplified base is 8 (not 12).

f) The simplified base is 8?

Yes; this was just proved.

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The outer and inner triangles are both equilateral and the circle touches all three sides of the outer triangle. If the area of
Brrunno [24]

Answer:

The area of outer Δ = 40 cm²

Step-by-step explanation:

∵ Area of the small triangle = 10 cm²

If we join the center of the circle with the 3 vertices of the inner Δ

These 3 segments are the radii of the circle

Now the inner triangle has 3 isosceles Δ their sides are r , r and s1 with vertex angle 120° ⇒ (360° ÷ 3 = 120°)

Where r is the radius of the circle and s1 is the side of the inner triangle

<em>By using cosine rule</em>

(s1)² = r² + r² - 2r²cos120 = r² + r² - 2r² (-0.5) = r² + r² + r² = 3r²

∴ s1 = r√3

∵ The radius of the circle ⊥ to the side of the outer Δ because the side of the outer Δ is a tangent to the circle

If we join a vertex of the outer Δ with the center of the circle

We will have a right angle triangle of two legs r and half s2 with angle 60° (120° ÷ 2 )between them ⇒ s2 is the side of outer Δ

∴ tan 60° = 1/2 (s2) ÷ r ⇒ √3 = 1/2 (s2) ÷ r = (s2)/2r

∴ s2 = 2r√3

∴ s2 : s1 = 2r√3 ÷ r√3 = 2 : 1

∴ The side of the outer Δ is double the side of the inner Δ

<em>By using similarity ratio</em>

A2/A1 = (s2/s1)² ⇒ A2 and A1 are the areas of outer and inner triangles

∴ A2 : A1 = (2/1)² = 4/1  

∴ A2 = 4 A1

∴ A2 = 4 × 10 = 40 cm²

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attashe74 [19]
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2x + 15y = 20     

</span><span>Multiply the 1st equation by -1:

-</span>2x - 3y = -20
2x + 15y = 20 
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       12y = 0, so y = 0.  Then the first equation becomes  2x + 3(0) = 20, or
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Solution is (10,0).
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3 years ago
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<span>The multiplicity of a zero of a polynomial function is how many times a particular number is a zero for a given polynomial.

For example, in the polynomial function f(x)=x(x+4)^2(x-2)^3, the zeros are 0 with a multiplicity of 1, -4 with a multiplicity of 2, and 2 with a multiplicity of 3.

Although this polynomial has only three zeros, we say that it has six zeros (or degree of 6) counting the <span>multiplicities.</span></span>
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312/2=156

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