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3 years ago

Find the range and the domain of the function below

Mathematics

1 answer:

Tasya [4]3 years ago

7 0

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

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Which expression is equivalent to (9x10z2)(x5y3)

MAVERICK [17]

Answer:

9x^15y^3z^2 is your answer

Step-by-step explanation:

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2 years ago

Find the product of (x-7)^2 and explain how it demonstrates the closure property of multiplication

iogann1982 [59]

A. x²-14x+49; is a polynomial

Step-by-step explanation:

(x-7)² can be written as (x-7)(x-7)

Expanding the expression

x(x-7)-7(x-7)

x²-7x-7x+49

x²-14x+49 ⇒⇒A quadratic function, which is a polynomial of degree 2

This function demonstrates the closer property of multiplication in that the change in order of multiplication does not change the product. This is called commutative property.

(x-7)(x-7)

-7(x-7)+x(x-7)

-7x+49+x²-7x

x²-14x+49

Learn More

Polynomials :brainly.com/question/9601478

Keywords : product, closure property of multiplication,

#LearnwithBrainly

4 0

2 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$