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3 years ago

How much canvas is needed to cover the bottom of a tent if the center pole is 7ft from the edge?

Mathematics

1 answer:

juin [17]3 years ago

8 0

The area (A) of a circle with radius r is given by

... A = π·r²

For a radius of 7 ft, the area is

... A = π·(7 ft)² = 49π ft² ≈ 154 ft²

The area of the bottom of a circular tent with radius 7 ft is about 154 ft².

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PLS HELP ME ASAP FOR 9) A and B (SHOW WORK!!) + LOTS OF POINTS!

statuscvo [17]

A.. 9.5*3.6=34.2 then 34.2/5.7= 6 so 6 is your answer

B. 16.2*5.7= 92.34 then 92.34/3.6= 25.65

4 0

3 years ago

Read 2 more answers

Sara jumps off a diving board that is 6 feet above the surface of the water her height above the surface of the water h in feet

Flauer [41]

Well when she hits the water h=0 so

2t^2-t-6=0 factor

2t^2-4t+3t-6=0

2t(t-2)+3(t-2)=0

(2t+3)(t-2)=0, since t>0

t=2 seconds...

However it should be noted that h(t)=-2t^2+t+6 implies that

v(t)=dh/dt=-4t+1 and

a(t)=dv/dt=-4

Since this is a free fall equation, this would suggest that the acceleration due to gravity is -4, whether that is supposed to be -4ft/s^2 or -4m/s^2, it certainly is not on this planet! :D If it is -4ft/s^2 it could be on the Moon or if it is -4m/s^2 it could be on Mars.

7 0

3 years ago

Rectangle A measures 9 inches by 3 inches. Rectangle B is a scaled copy of rectangle A.Select all of the measurements pairs that

umka2103 [35]

Answer:

3 inches by 1 inches or 6 inches by 2 inches. Both answera are correct

Step-by-step explanation:

just an equivalent ratio

hope this helps can I get brainliest

7 0

3 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

#SPJ4

6 0

1 year ago

Consider the following linear equation.y = -1-Step 1 of 2: Determine the slope and the y-intercept (entered as an ordered pair)

Dmitriy789 [7]

Recall that a slope-intercept form of a linear equation is

$\begin{gathered} y=mx+b \\ \text{where} \\ m\text{ is the slope} \\ b\text{ is the y-intercept} \end{gathered}$

Rearrange the given so that it follows that slope intercept form

$\begin{gathered} y=-1-\frac{2}{5}x\Longrightarrow y=-\frac{2}{5}x-1 \\ \\ \text{Now that it is in the slope intercept form} \\ y=-\frac{2}{5}x-1 \\ \text{we can determine that the slope is } \\ m=-\frac{2}{5} \\ \text{and the y-intercept is} \\ b=-1 \end{gathered}$

The y-intercept is the value for which the linear function crosses the y-axis, this is when the value of x is equal to zero. The ordered pair therefore of the y-intercept is (0,-1).

8 0

1 year ago