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Nina [5.8K]
2 years ago
13

25-20x+4x² heh help pls and hurry :((

Mathematics
2 answers:
ipn [44]2 years ago
8 0

Answer:100 POINTS! (50 split up) PLSSSSSSSSSS HELP!!! I REALLY NEED IT SO MUCH :( I PROMICE TO GIVE BRAINLIEST AND FIVE STARS AND A THANKS! 6TH GRADE SCIENCE! i used all my points for this!

You measure the mass of an apple using a balance. You get these three measurements:

100

yulyashka [42]2 years ago
6 0
25- 20x + 4x^2
=4x^2 - 20x + 25
=4x^2 - x(10+10) +25
=4x^2 - 10x - 10x + 25
=2x(2x - 5) - 5( 2x - 5)
=(2x - 5)(2x - 5)
=(2x - 5)^2
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The surface area of this triangular pyramid, whose base and lateral faces are congruent equilateral triangles, is __ square inch
DedPeter [7]

Answer:

The surface area is equal to 390\ in^{2}

Step-by-step explanation:

The surface area of the triangular pyramid is equal to the area of its four triangular faces

so

In this problem

SA=4[\frac{1}{2}(b)(h)]

we have

b=15\ in

h=13\ in

substitute the values

SA=4[\frac{1}{2}(15)(13)]=390\ in^{2}

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A worker at an orchard measures the mass of different numbers of apples. The scatter plot below shows the relationship between t
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0.12

Step-by-step explanation:

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Look at the following pairs of numbers (x, y).
katrin2010 [14]

Step-by-step explanation:

Taking the first coordinate point (3,16.5)

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3 years ago
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A train is spotted 10 miles south and 8 miles west of an observer at 2:00 pm. At 3:00 pm the train is spotted 5 miles north and
kodGreya [7K]

Answer:

a. The distance the train travelled in the first hour is approximately 28.3 miles

b. The location of the train at 5:00 p.m. is 53 miles east, and 46 miles west

c. The location of the train at any given time by the function, f(t) = (-8 + 24·t, -10 + 15·t)

d. The train does not collide with the cyclist when the bike goes over the train tracks

Step-by-step explanation:

a. The given information on the train's motion are;

The location south the train is spotted = 10 miles south and 8 miles west

The time the observer spotted the train = 2:00 pm

The location the train is spotted at 3:00 p.m. = 5 miles north and 16 miles east

Therefore, the difference between the two times the train was spotted, t = 3:00 p.m. - 2:00 p.m. = 1 hour

Making use of the coordinate plane for the two locations the train was spotted, we have;

The initial location of the train = (-10, -8)

The final location of the train = (5, 16)

Therefore the distance the train travelled in the first hour is given by the formula for finding the distance, 'd', between two points, (x₁, y₁) and (x₂, y₂) as follows;

d = \sqrt{\left (x_{2}-x_{1}  \right )^{2}+\left (y_{2}-y_{1}  \right )^{2}}

Therefore;

d = \sqrt{\left (5-(-10)  \right )^{2}+\left (16-(-8)  \right )^{2}} = 3 \cdot\sqrt{89}

The distance the train travelled in the first hour, d = 3·√89 ≈ 28.3 miles

b. The speed of the train, v = (Distance travelled by the train)/Time

∴ v ≈ 28.3 miles/(1 hour) = 28.3 miles per hour

The speed of the train in the first hour, v ≈ 28.3 mph

The direction of the train, θ, is given by the arctangent of the slope, 'm', of the path of the train;

\therefore The  \  slope  \  of \ the \  path  \ of \  the \  train, \, m =tan(\theta) = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

∴ m = tan(θ) = (5 - (-10))/(16 - (-8)) = 0.625

c. Distance = Velocity × Time

At 5:00 p.m., we have;

The time difference, Δt = 5:00 p.m. - 3:00 p.m. = 2 hours

The distance, d₁ = (28.3 mph × 2 hours = 56.6 miles

Using trigonometry, we have the horizonal distance travelled, 'Δx', in the 2 hours is given as follows;

Δx = d₁ × cos(θ)

∴ Δx = 56.6 × cos(arctan(0.625)) ≈ 48

The increase in the horizontal position of the train, relative to the point (5, 16), Δx ≈ 48 miles

The vertical distance increase in the two hours, Δy is given as follows;

Δy = 56.6 × sin(arctan(0.625)) ≈ 30

The increase in the vertical position of the train, relative to the point (5, 16), Δy ≈ 30 miles miles

Therefore; the location of the train at 5:00 p.m. = ((5 + 48), (16 + 30)) = (53, 46)

The location of the train at 5:00 p.m. = 53 miles east, and 46 miles west

c. The function, 'f', that would give the train's position at time-t is given as follows;

The P = f(28.3·t, θ)

Where;

28.3·t = √(x² + y²)

θ = arctan(y/x)

Parametric equations

y - 5 = 0.625·(x - 16)

∴ y = 0.625·x - 10 + 5

The equation of the train's track is therefore, presented as follows;

y = 0.625·x - 5

d = 28.3·t

The y-component of the velocity, v_y = 3*√89 mph × sin(arctan(0.625)) = 15 mph

Therefore, we have;

y = -10 + 15·t

The x-component of the velocity, vₓ = 3*√89 mph × cos(arctan(0.625)) = 24 mph

Therefore, we have;

x = -8 + 24·t

The location of the train at any given time, 't', f(t) = (-8 + 24·t, -10 + 15·t)

d. The speed of the cyclist next to the observer at 2:00 p.m., v = 10 mph

The distance of the cyclist from the track = The x-intercept = 5/0.625 = 8

The distance of the cyclist from the track = 8 miles

The time it would take the cyclist to react the track, t = 8 miles/10 mph = 0.8 hours

The location of the train in 0.8 hours, is f(0.8) = (-8 + 24×0.8, -10 + 15×0.8)

∴ f(0.8) = (11.2, 2)

At the time the cyclist is at the track along the east-west axis, at the point (8, 0), the train is at the point (11.2, 2) therefore, the train does not collide with the cyclist when the bike goes over the train tracks.

8 0
2 years ago
Line A: x + y = 6 Line B: x + y = 4 Which statement is true about the solution to the set of equations? a There are infinitely m
vovangra [49]

Answer:

55

Step-by-step explanation:

55 is the answer I hope this helps

3 0
3 years ago
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