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2 years ago

14

A 10-foot ladder is leaning against a wall. The base of the ladder is 2 feet from the base of the building. How far up the build

ing does the ladder fall?

1 answer:

Pavel [41]2 years ago

5 0

10x2 = 20

the building from the ladder is 20ft

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Aaron bought 7/8 of a yard of fabric and it cost him $4.10. What does a full yard cost?

gayaneshka [121]

Divide 4.10 by 7 to get about 0.59 (round up because you can't get part of a penny) and that gives you 1/8 of a yard's price. Then multiply that by 8 to get a full yard for $4.72.

7 0

3 years ago

A scientist planted seeds in 4 sections of soil for an experiment. Not all of the

kakasveta [241]

Answer:

I think you would multiply 20 × 4 and the answer would be 80 that the scientist counted the number of plants in each 4 sections after 20 days.

4 0

2 years ago

Graph the equation <br><br>2x + 3y = 9

jeka57 [31]

$2x + 3y = 9\\3y = -2x + 9\\y = -\frac{2}{3}x + 3$

Look at the graph picture I attached

$y = mx + c$

c is the intercept, which is why the graph crosses the y-axis at +3

to find where it crosses the x-axis, we make y = 0

$y = 0\\\to -\frac{2}{3}x + 3 = 0\\-\frac{2}{3} = -3\\\frac{2}{3}x = 3\\2x = 9\\x = \frac{9}{2}\ \ (=4.5)$

8 0

2 years ago

If the radius is 18 cm and the angle at centre of circle is 120° then the

PolarNik [594]

The area of a sector is 339.336 cm²

The area of a sector is defined to be as the region of space bounded by a boundary from the center of the circle.

It can be determined by using the formula:

$\mathbf{A = \dfrac{\theta }{360}\times \pi r^2 }$

Given that;

Radius r = 18 cm
Angle θ = 120°

Then. the area of the sector can be calculated as:

$\mathbf{A = \dfrac{120}{360}\times \pi (18)^2 }$

$\mathbf{A = \dfrac{1}{3}\times 3.142 \times 324 }$

A = 339.336 cm²

Learn more about the area of a sector here:

brainly.com/question/7512468?referrer=searchResults

4 0

2 years ago

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

2 years ago