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3 years ago

7

Now let the figure show a scale drawing of a park. The scale is 1 unit : 25 meters. What is the horizontal distance across the p

ark?
A: 14 m
B: 225 m
C: 350 m

1 answer:

insens350 [35]3 years ago

7 0

The question is missing the figure. So, the figure is attached below.

Answer:

C. 350 m

Step-by-step explanation:

Given:

Scale is given as:

1 unit : 25 meters

This means that 1 unit on the grid is equivalent to 25 meters in actual.

Now, from the figure, the horizontal distance across the park is 14 units.

1 unit = 25 meter

Now, we can find the actual distance using unitary method and thus multiplying 25 and 14 to get the actual distance across the park horizontally.

∴ 14 units = $25\times 14=350$ meters.

Therefore, the horizontal distance across the park is 350 m in actual.

So, the correct option is option C.

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Vince has a rectangular rug in his room with an area of 10 ft the length of the rug is 18 inches longer than the width what coul

aev [14]

The length of the rug is 4 ft.

The width of the rug is 2.5 ft.

Explanation:

The area of the rug is 10 ft.

The length of the rug be l.

Let us convert the inches to feet.

Thus, $18 inches = 1.5 ft$

Thus, the length of the rug is $l=1.5+w$

Let the width of the rug be w.

Substituting these values in the formula of area of the rectangle, we get,

$A=length\times width$

$10=(1.5+w)(w)\\10=1.5w+w^2\\w^2+1.5w-10=0$

Solving the expression using the quadratic formula,

$w=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Substituting the values, we have,

$$w=\frac{-15 \pm \sqrt{15^{2}-4 \cdot 10(-100)}}{2 \cdot 10}\\$

$w=\frac{-15 \pm \sqrt{4225}}{20}$

$w=\frac{-15 \pm 65}{2 0}$

Thus,

$w=\frac{-15 + 65}{2 0}\\w=\frac{50}{20} \\w=2.5$ and $w=\frac{-15 - 65}{2 0}\\w=\frac{-80}{20} \\w=-4$

Since, the value of w cannot be negative, the value of w is 2.5ft

Thus, the width of the rug is 2.5ft

Substituting $w=2.5$ in $l=1.5+w$ , we get,

$l=1.5+2.5\\l=4$

Thus, the length of the rug is 4 ft.

4 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

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$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

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3 years ago

Which of the following is the graph of f(x) = 3 sin (2x)? (2 points)

TEA [102]

Answer:

Step-by-step explanation:

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3 years ago

While in transit, the antitoxin had to stay between 36ºF and 46ºF. The dogsled teams stopped along the route often to warm the a

eimsori [14]

Range is the difference between the largest and the smallest numbers in a given contest

Hence the range
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2 years ago

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mash [69]

Answer:

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Step-by-step explanation:

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= 67.7

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3 years ago