All categories

4 years ago

Use a composite figure to estimate the area of the figure. The grid has squares with side lengths of 1.5 cm.

Mathematics

1 answer:

Irina-Kira [14]4 years ago

4 0

Answer:

about $49.1\ cm^{2}$

Step-by-step explanation:

The area of the composite figure is approximate to the area of the rectangle plus the area of the semicircle at the top of the rectangle plus the semicircle at the right of the rectangle

Find the area of the rectangle

$A=(3*1.5)(4*1.5)=27\ cm^{2}$

Find the area of the semicircle at the top of the rectangle

$A=\frac{1}{2}\pi (2*1.5)^{2}=4.5 \pi\ cm^{2}$

Find the area of the semicircle at the right of the rectangle

$A=\frac{1}{2}\pi (1.5*1.5)^{2}=2.53 \pi\ cm^{2}$

The area of the composite figure is

$27\ cm^{2}+4.5 \pi\ cm^{2}+2.53 \pi\ cm^{2}=(27+7.03 \pi)\ cm^{2}$

use $\pi=3.14$

$(27+7.03(3.14))=49.1\ cm^{2}$

You might be interested in

What is the color for 1.99 for goodwill tomorrow?

Ilya [14]

Depends wut state u in

3 0

3 years ago

Esther is participating in a walk-a-thon. She will raise 5 dollars for her charity for each lap she walks.

EastWind [94]

Answer:

answer d

Step-by-step explanation:

I took the assignment and got it right :)

6 0

3 years ago

Read 2 more answers

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago

Two nonadjacent angles formed by two intersecting lines are

Sophie [7]

Two vertical angles are formed. These angles are also equal to each other.

6 0

3 years ago

Please answer this question now

Alex787 [66]

Answer:

541.7 (m2)

Step-by-step explanation:

Applying the sine theorem:

WV/sin(X) = XV/sin(W)

=> WV = XV*sin(X)/sin(W) = 37*sin(50)/sin(63) = 31.81

Angle V = 180 - X - W = 180 - 50 - 63 = 67

Denote WH is a height of the triangle VWX, H lies on XV

=> WH = WV*sin(V) = 31.81*sin(67) = 29.28

=> Area of triangle VWX is calculated by:

S = side*height/2 = XV*WH/2 = 37*29.28/2 = 541.7 (m2)

8 0

3 years ago