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butalik [34]
3 years ago
11

Use the graph below to find f(1)

Mathematics
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

I dont understand sorry

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a,c and possibly d

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8 0
3 years ago
A 13​-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 5 ft from the​ house, the
lord [1]

Answer:

a. -5 ft per sec

b. 59.5 ft² per sec

Step-by-step explanation:

a. Suppose l represents the ladder, x represents the height of the top of ladder from the base of house and y represents the distance of base of ladder from the base of house.

By the Pythagoras theorem,

l^2 = x^2 + y^2------(1)

Differentiating with respect to t ( time ),

2l\frac{dl}{dt} = 2x \frac{dx}{dt}+2y\frac{dy}{dt}

But, height of ladder, l = 13 ft = constant,

\implies \frac{dl}{dt}=0

\implies 0= 2x \frac{dx}{dt}+2y\frac{dy}{dt}-----(2)

We have,

y = 5, \frac{dy}{dt}=12\text{ ft per sec},

Again by equation (1),

13^2 = x^2 + 5^2\implies 169 = x^2 + 25\implies 169 - 25 = x^2\implies 144 = x^2\implies x = 12\text{ ft}

From equation (2),

0 = 2(12) \frac{dx}{dt} + 2(12)(5)

0= 24\frac{dx}{dt}+120

-120 =24\frac{dx}{dt}

\implies \frac{dx}{dt}=-\frac{120}{24}=-5\text{ ft per sec}

Hence, the rate of change of the height of the top of the​ ladder is -5 ft per sec.

b. Now area of the triangle = 1/2 × base × height

\implies A = \frac{1}{2}\times y\times x

Differentiating with r. t. t,

\frac{dA}{dt}=\frac{1}{2}(y\frac{dx}{dt}+x\frac{dy}{dt})

= \frac{1}{2}(5(-5) + 12(12))

=\frac{1}{2}(-25 + 144)

=\frac{119}{2}

= 59.5 ft² per sec

Hence, area of the triangle is changing with the rate of 59.5 ft per sec.

7 0
4 years ago
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