8/15 + 2/5 pls solve this for me someone pls

There’s also D D: 4m on top and 4m on the side

Wewaii [24]

B hope this helped have a nice day! =D

Oh yes 4x5 is another answer so yep god I forgot that sorry!

4 0

3 years ago

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#1. Use substitution to solve the system of equations

Bond [772]

Answer:

y= −2/3x+−20/3

heart if it helped you brain if im cool

3 0

3 years ago

of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10

JulsSmile [24]

Answer:

y = -0.11x^2 + 1.111x

y = 28 ft .... Height at center

Step-by-step explanation:

Given:-

- The span of the arc is = 100 ft

- The height of the arch is 40 ft at 10 ft from center.

Find:-

- The equation of parabolic arch and the height of the arch at center.

Solution:-

- We will take the height y as a function of width x of the parabolic arch. The general equation of the arch is such that it passes through origin. The equation is given in the form as:

y = f(x) = ax^2 + bx

Where,

a, b, and c are constants to be determined.

- We will use the condition i.e the span of entire arch is 100 ft. So we could say that y = 0 for x = 100 ft. Then we have:

0 = f(100) = a(100)^2 + b(100) ..... 1

- Using second condition i.e y = 10 ft at 40 ft from center. Since, due to symmetry we know that center lies at x = 50 ft. Then y = 10 ft at x = 10 ft. The condition can be expressed in the form:

10 = f(10) = a(10)^2 + b(10) ..... 2

- Solving the 2 Equations simultaneously, we have:

0 = a(100)^2 + b(100)

10*10 = a*10*(10)^2 + b(10)*10

100 = a(10)^3 + b(100)

- Subtract both equations:

100 = a*(10^3 - 100^2)

a = 100 / ( 1000 - 10000)

a = -0.11

- Then using a = -0.11 evaluate b:

-1.11 + 10b = 10

b = 11.11 / 10 = 1.111

- The equation of the parabola is:

y = -0.11x^2 + 1.111x

-The height of the arch at center where x = 50 ft.

y = -0.11(50)^2 + 1.111(50)

y = -27.5 + 55.5

y = 28 ft

- The height of the parabolic arch at center is given as y = 28 ft.

4 0

3 years ago

If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi

Finger [1]

ANSWER:

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

SOLUTION:

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = $\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}$ --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

$\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16$

$34 \times 1+\frac{1}{2} \times a \times 1^{2}=18$

$34+\frac{a}{2}=18$

$\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}$

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0

3 years ago

If cos theta is less than 0 and cot theta is greater than 0, then the terminal point determined by theta is in:

Keith_Richards [23]

There exist an abbreviation that ALL - S - T - C where all trigonometric functions in first quandrant are positive. S, T, and C are the first letters of the trigonometric functions that are positive in quadrant 2, 3, and 4, respectively. This also means that in the same quadrant, their reciprocals are also positive. For the given above, it is in Quadrant 3 where T is positive and cosine is negative.

5 0

3 years ago

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