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3 years ago

11

A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = -16t2 + 704t

. After how many seconds does the projectile take to reach its maximum height?

1 answer:

Alex787 [66]3 years ago

4 0

You need to find the x-coordinate (h) of the vertex of the parabola.

$-16t^2+704t=0\\
h=-\dfrac{b}{2a}=-\dfrac{704}{2\cdot(-16)}=22
$

After 22 seconds.

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What is pythagoras theorem

Alborosie

Answer:

hypotenuse²=base²x height²

Step-by-step explanation:

It is used to find the third side of a triangle.

Hope it helps :)

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4 years ago

3⋅ 3/5 (3/5 is a fraction)

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Well its a simplest form of zero and 6 tenths and if u divide it by 2 youll get 3/5 as a fraction

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4 years ago

A fruit company delivers its fruit in two types of boxes: large and small. A delivery of

Tema [17]

Answer:

Small box weighs 13.75 kg & large box weighs 15.75 kg

Step-by-step explanation:

We can write 2 simultaneous equation and solve for weight of each box.

<em>Let weight of large box be l and small box be s.</em>

<em />

"<u>3 large boxes and 5 small boxes has a total weight of 116 kilograms</u>":

$3l+5s=116$

and

"<u>9 large boxes and 7 small boxes has a total weight of 238 kilograms</u>":

$9l+7s=238$

<em>Now we can solve for l in the 1st equation and put it into 2nd equation and get s:</em>

<em> $3l+5s=116\\3l=116-5s\\l=\frac{116-5s}{3}$ </em>

<em>now,</em>

<em> $9l+7s=238\\9(\frac{116-5s}{3})+7s=238\\3(116-5s)+7s=238\\348-15s+7s=238\\348-238=15s-7s\\110=8s\\s=\frac{110}{8}=13.75$ </em>

<em />

<em>now we plug in 13.75 into s into equation of l to find s:</em>

<em> $l=\frac{116-5s}{3}\\l=\frac{116-5(13.75)}{3}\\l=15.75$ </em>

5 0

3 years ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

What is the equation of a line that passes through (4, 7) and has a slope of −14?

deff fn [24]

Slope = -1/4
(4, 7)

y - 7 = -1/4 (x - 4)
y - 7 = -1/4x + 1
y = -1/4 + 8

answer is C

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3 years ago