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murzikaleks [220]
3 years ago
15

Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, aft

er t seconds that Amir has thrown the ball is given by:
H (t) = -(t-2)^2+9
Mathematics
1 answer:
telo118 [61]3 years ago
8 0

Answer:

Time t = 2 seconds

It will reach the maximum height after 2 seconds

Completed question;

Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, after t seconds that Amir has thrown the ball is given by:

H (t) = -(t-2)^2+9

many seconds after being thrown will the ball reach its maximum height?

Step-by-step explanation:

The equation of the height!

h(t) = -(t-2)^2 + 9 = -(t^2 -4t +4) + 9

h(t) = -t^2 +4t -4+9

h(t) = -t^2 + 4t +5

The maximum height is at dh/dt = 0

dh/dt = -2t +4 = 0

2t = 4

t = 4/2 = 2

Time t = 2 seconds

It will reach the maximum height after 2 seconds

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If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?
eduard

Answer:

cos(\alpha+\beta)=-\frac{84}{85}

Step-by-step explanation:

we know that

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

Remember the identity

cos^{2} (x)+sin^2(x)=1

step 1

Find the value of sin(\alpha)

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

cos(\alpha)=-\frac{8}{17}

Find the value of sine

cos^{2} (\alpha)+sin^2(\alpha)=1

substitute

(-\frac{8}{17})^{2}+sin^2(\alpha)=1

sin^2(\alpha)=1-\frac{64}{289}

sin^2(\alpha)=\frac{225}{289}

sin(\alpha)=-\frac{15}{17}

step 2

Find the value of cos(\beta)

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

sin(\beta)=-\frac{4}{5}

Find the value of cosine

cos^{2} (\beta)+sin^2(\beta)=1

substitute

(-\frac{4}{5})^{2}+cos^2(\beta)=1

cos^2(\beta)=1-\frac{16}{25}

cos^2(\beta)=\frac{9}{25}

cos(\beta)=\frac{3}{5}

step 3

Find cos⁡ (α + β)

cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)

we have

cos(\alpha)=-\frac{8}{17}

sin(\alpha)=-\frac{15}{17}

sin(\beta)=-\frac{4}{5}

cos(\beta)=\frac{3}{5}

substitute

cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})

cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}

cos(\alpha+\beta)=-\frac{84}{85}

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