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Naddika [18.5K]
3 years ago
11

A 7 foot long board is propped against the wall of a house. The board forms a 60 angle with the ground. How far is the base of t

he board from the wall?
A. 3.5 ft

B. 3 ft

C. 2.5 ft

D. 4 ft
Mathematics
2 answers:
Citrus2011 [14]3 years ago
5 0
This is like a 30-60-90 triangle with a hypotenuse of 7. Since the relationship of the base to the hypotenuse in a 30-60-90 triangle is 1:2, the answer is 7/2=3.5 ft. Hope this helps!
Bingel [31]3 years ago
3 0
3.5 ft 
thats ur answer your welcome

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The coordinates of a triangle's vertices are A(2, 4), B(3, 1), and C(5, 5). The triangle undergoes a translation 4 units to the
AlekseyPX
South west

Shifting 4 units left (decreasing in x-axis by 4 units)
Shifting 5 units down (decreasing in y-axis by 5 units)

Apply to every vertices coordinates(x - 4, y - 5)

New coordinates

A(-2, -1)
B(-1, -4)
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If the slope of one of these parallel lines is 1.25, the slope of the other line is ___.
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A student ran 100 m dash in 15.4 seconds. What was the student speed in miles per hour
amid [387]

15.4 seconds = 0.00428 hours and 100 meters = 0.062 miles soo the student ran approximately 14.5 mph

8 0
3 years ago
b) If parametric equations of a flow line are x = x(t), y = y(t), explain why these functions satisfy the differential equations
sineoko [7]

Answer:

The equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1.

Step-by-step explanation:

The pathline equation for a vector field is given by F(x,y) = xî - yj

The velocity vector field for the streamline of the flow is given by

V(x, y) = (dx/dt)î + (dy/dt)j

From the question, it is given that

(dx/dt) = x

(dy/dt) = -y

Hence, the velocity vector field for the streamline of the flow in question is

V(x, y) = xî - yj

which coincides with the pathline vector field of the flow.

The only time the pathline and streamline vector field coincide and have the same equation is when the flow is a steady state flow.

That is, the properties of the fluid flowing isn't changing with time!

Hence, this flow is a steady state flow!

We're told to solve the differential equation.

(dx/dt) = x

(dy/dt) = -y

but

(dy/dx) = (dy/dt) × (dt/dx)

(dy/dx) = -y/x

(dy/y) = -(dx/x)

∫(dy/y) = -∫ (dx/x)

In y = - In x + c

where c is the constant of integration

In y + In x = c

In (xy) = c

Inserting the values of (x, y) given in the question,

In (-1 × -1) = c

In 1 = c

0 = c

c = 0

In y + In x = 0

In (yx) = 0

xy = e⁰ = 1

xy = 1

So, the equation of the the flow line that passes through the point (x, y) = (−1, −1) is

In y + In x = 0 or in another form, xy = 1

Hope this Helps!!!

4 0
3 years ago
Petra was given 12 yards of fabric to make costumes for the school play if each costume requires 1 1/2 yards how many costumes c
solmaris [256]

Petra can make 8 costumes with 12 yards of fabric

<h3><u>Solution:</u></h3>

Given that Petra was given 12 yards of fabric to make costumes for the school play

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<em><u>To find: number of costume Petra can make</u></em>

The number of costume Petra can make is calculated by dividing the total fabric she has by fabric needed for one costume

\text {number of costume Petra can make }=\frac{\text {total fabric she has }}{\text {fabric needed for one costume }}

Total fabric she has = 12 yards

Fabric needed for 1 costume = 1\frac{1}{2} = \frac{2 \times 1 + 1}{2} = \frac{3}{2} yards

Therefore,

\text {number of costume Petra can make }=\frac{12}{3 / 2}=\frac{12}{3} \times 2=8

Thus petra can make 8 costumes with 12 yards of fabric

7 0
3 years ago
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