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3 years ago

11

Write an expression to represent "triple the quotient of a number and three."

1 answer:

Furkat [3]3 years ago

4 0

Answer:

3n/3 or 3n÷3

Step-by-step explanation:

since it says triple the quotient of a <u>number</u>, we can use the number n.

To find the quotient means to divide and it's going to be by that numer.....n....... trippled (3)

put them all together and you get 3n/3

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If a triangle has side lengths 5 cm, 18<br> cm, and 12 cm, is it a right triangle?

Dimas [21]

Answer:

No, it's not a right triangle.

Step-by-step explanation:

1. The Pythagorean Theorem states that the area of the square whose side is the hypotenuse (longest side of a right triangle) is equal to the sum of the area of the squares from the two other sides.

2. The longest side in this "triangle" has 18 centimeters, and the other 2 sides are 5 and 12 centimeters.

3. Let's apply the Pythagorean Theorem to see if 5^2 + 12^2 = 18^2

$5^2 + 12^2 = 18^2$
$25 + 144 = 324$
$169 \neq 324$

4. Okay, so the sum of the area of the squares from the two other sides is not equal to the hypotenuse squared, therefore, it's not a right triangle.

(Plus, it's not even a triangle because 7 < x < 17, and the third "side", 18, is not between 7 and 17.)

5 0

3 years ago

Save me the headache

maxonik [38]

$(9\sin2x+9\cos2x)^2=81$

Taking the square root of both sides gives two possible cases,

$9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1$

or

$9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1$

Recall that

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

If $\alpha=2x$ and $\beta=\dfrac\pi4$ , we have

$\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}$

so in the equations above, we can write

$\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1$

Then in the first case,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi$

(where $n$ is any integer)

$\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi$

$\implies x=n\pi\text{ or }\dfrac\pi4+n\pi$

and in the second,

$\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}$

$\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi$

$\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi$

$\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi$

Then the solutions that fall in the interval $[0,2\pi)$ are

$x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4$

5 0

3 years ago

Read 2 more answers

Use a net to find the surface area of the cylinder use 3.14 for pi

telo118 [61]

Answer:

3.14

Step-by-step explanation:

7 0

3 years ago

Show that the equation x^3 + x -3 =0 has a solution between 1 and 2. That is the only information given

valentina_108 [34]

Answer:

Step-by-step explanation:

Given that:

$f(x) = x^3 + x - 3 = 0$

Since the given function is equal to zero, then the function :

$f(x) = x^3 + x - 3$

where;

$x = 1$ and

$x = 2$

$f(x) = (1)^3 + (1) - 3 \\ \\ f(x) = 1 + 1 - 3 \\ \\f(x) = -1 < 0$

$f(2) = x^3 + x - 3 \\ \\f(2) = (2)^3 + 3 - 3 \\ \\f(2) = 8 + 3 - 3 \\ \\f(2) = 11 - 3 \\ \\f(2) = 8 > 0 \\ \\$

Thus; f(1) < 0 and f(2) > 0

8 0

3 years ago

Is 7/9 squared Rational or irrational?

Karo-lina-s [1.5K]

the answer is irrational

8 0

3 years ago