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3 years ago

Find the value of x. Round to the nearest degree as needed.

Mathematics

1 answer:

Ainat [17]3 years ago

5 0

9514 1404 393

Answer:

x° = 59°

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the trig relations. The one applicable here is ...

Tan = Opposite/Adjacent

tan(x°) = 5/3

x° = arctan(5/3) ≈ 59.04°

The value of x is about 59.

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3 years ago

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Answer:

2.0 seconds

Step-by-step explanation:

Given quadratic functions:

$\begin{cases}y=-7x^2+26x+3\\y=-6x^2 +23x+5\end{cases}$

To find the time, in seconds, that the balloons collided at the highest point, substitute one equation into the other equation and rearrange to equal zero:

$\begin{aligned}-6x^2+23x+5 & = -7x^2+26x+3\\-6x^2+23x+5+7x^2 & = -7x^2+26x+3+7x^2\\x^2+23x+5 & = 26x+3\\x^2+23x+5-26x & = 26x+3-26x\\x^2-3x+5& = 3\\x^2-3x+5-3& = 3-3\\x^2-3x+2& = 0\end{aligned}$

Factor the quadratic:

$\begin{aligned}x^2-3x+2 & = 0\\x^2-2x-x+2 & = 0\\x(x-2)-1(x-2) & = 0\\(x-1)(x-2) & = 0\\\end{aligned}$

Apply the zero-product property to solve for x:

$\implies (x-1)=0 \implies x=1$

$\implies (x-2)=0 \implies x=2$

Therefore, the balloons collided at 1 second and 2 seconds.

To find at which time the highest point of collision occured, substitute both values of x into one of the functions:

$f(1)=-6(1)^2 +23(1)+5=22$

$f(2)=-6(2)^2 +23(2)+5=27$

Therefore, the time, in seconds, that the balloons collided at the highest point is 2.0 seconds.

Learn more about quadratic systems of equations here:

brainly.com/question/27930827

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1 year ago

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Write the expression as either the sine, cosine, or tangent of a single angle.

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There exists a trigonometric identity which states that,
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