Ask question

Ask question

All categories

3 years ago

7

The base of a triangle is shrinking at a rate of 1cm/min and the height if the triangle is increasing at a rate of 5cm/min. Find

the rate at which the area of the triangle changes when the height is 22 cm and the base is 10 cm.

1 answer:

lapo4ka [179]3 years ago

3 0

Answer: Rate of change of Area of the triangle = 2.75cm~/min

Step-by-step explanation:

If the present height of the triangle is 22cm and the height increases by 5cm/min. It means the height was 2cm four minutes earlier: 5cm for 1st min + 5cm for 2nd min + 5cm for 3rd min + 5cm for 4th min = 20cm after 4min.

Therefore height equals 22 - 20 = 2cm four minutes earlier.

Using the same four minutes for the base. Since the base shrinks at 1cm/min, for the base to shrink to 10cm it means it was 14cm four minutes earlier.

The Area of a triangle is given by

A = 1/2 × base × height

Therefore Area= 1/2× 10×22 = 110cm~

Since it took four minutes to get to this area, the rate of change of the area will be 110÷4 = 2.75cm~/min

You might be interested in

Which of the following options have the same value as 75%, percent of 96?

dexar [7]

The choices should be B, C, and E. Hope this helps

6 0

3 years ago

7 watermelons cost $8.96.Which equation would help determine the cost of the 5 watermelons?

postnew [5]

Answer: ($8.96 / 7) * 5 or $6.40

Step-by-step explanation: You need to find the price for one watermelon then multiply it by 5 to find the price of 5.

4 0

3 years ago

Read 2 more answers

1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Sveta_85 [38]

Answer:

1. $-5x+3y+44=0$

2. $2x+y-2=0$

3. $2x+y-4=0$

Step-by-step explanation:

Standard form of a line is $Ax+By+C=0$ .

If a line passing through two points then the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope, i.e., $m=\dfrac{y_2-y_1}{x_2-x_1}$ .

1.

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

$y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)$

$y+3=\dfrac{-5}{-3}(x-7)$

$y+3=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$-5x+3y+9+35=0$

$-5x+3y+44=0$

Therefore, the required equation is $-5x+3y+44=0$ .

2.

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to $2 x + y =-5$ .

Slope of the line : $m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2$

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

$y-2=-2(x-0)$

$y-2=-2x$

$2x+y-2=0$

Therefore, the required equation is $2x+y-2=0$ .

3.

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to $2x + y =-5$ .

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

$y-0=-2(x-2)$

$y=-2x+4$

$2x+y-4=0$

Therefore, the required equation is $2x+y-4=0$ .

5 0

3 years ago

C represents the commission on her sales.

vlabodo [156]

First simplify the expression.
1/3(1,650+0.15c)
550+0.05c
Answer is A
Hope this helps!

3 0

3 years ago

What is the rate of change of the volume of the cylinder at that instance (in cubic kilometers per second)?

Kay [80]

Answer:

hello your question is incomplete attached below is the complete question

answer : -2400 π

Step-by-step explanation:

dr/dt = -12 km/s

h = 2.5 km

r = 40 km at some instant

hence the rate of change of Volume of cylinder at that instance

= dv / dt = $d(\pi r^h ) / dt$

= $\pi h\frac{d(r^2)}{dt}$ = $-\pi h *2r *dr/dt$

hence ; dv / dt = -2400 π

3 0

3 years ago