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2 years ago

11

Given: ΔXYZ

1 answer:

castortr0y [4]2 years ago

8 0

28 / 2 = 14

Answer: 14 cm

In any triangle, the mid segment has half the length of the side to which it is parallel.

Hope this helps.

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F(x) = x^2 -7x +14, g(x) = 4x -9 <br><br> Find (g ° f)(x)

adelina 88 [10]

Answer:

Set up the composite result function.

g(F(x)) Evaluate g(x2−7x+14)

by substituting in the value of F into g.

g(x2−7x+14)=4(x2−7x+14)−9

Simplify each term.

g(x2−7x+14)=4x2−28x+56−9

Subtract 9 from 56.

g(x2−7x+14)=4x2−28x+47

Step-by-step explanation:

5 0

2 years ago

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f

zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time $t$

At $t=0, the weight of the bucket is 6+36=42 \mathrm{lb}$

After $t$ seconds, the weight of the bucket is $42-0.15 t \mathrm{lb}$

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after $t$ seconds is $x=1.5 t$

Since the well is 80 ft deep, the time to pull out the bucket is $\frac{80}{2}=40 \mathrm{~s}$

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

$W=\int_{t=0}^{t=36} F d x$

Use $x=1.5 t$ and $F=42-0.15 t$

$W=\int_{0}^{36}(42-0.15 t)(1.5 d t)$

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

$W=3200 \mathrm{ft} \cdot \mathrm{lb}$

4 0

2 years ago

Write a letter to the student council that describes

zhenek [66]

Answer:

its b on ege

Step-by-step explanation:

5 0

2 years ago

Hiii can someone pls pls pls help asap

AlekseyPX

Answer:

k = 3.5

Step-by-step explanation:

You need to divide all the numbers.

21/6 = 3.5

24.5/7 = 3.5

28/8 = 3.5

31.5/9 = 3.5

So, the answer is 3.5.

8 0

2 years ago

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the re

alukav5142 [94]

Answer:

Step-by-step explanation:

$\text{Given that:}$

$y = 1+ sec(x) \ \ y =3$

$\text{we draw the graph and the curves intersect at:}$

$x = - \dfrac{\pi}{3} \ and \ x = \dfrac{\pi}{3}$

$\text{Applying washer method;}$

$f(x) _{outer} - g(x) _{inner} --- (1)$

$V= \int ^b_a A(x) \ dx --- (2)$

$\text{outer radius = 3 - 1 = 2}$

$\text{inner radius =}$ $( 1 + sec(x) ) - 1 = sec (x)$

$A(x) = \pi ((2)^2 -(sec(x)^2) \\ \\ A(x) = \pi (4 - sec^2 (x)) ---- (3)$

$\text{The volume V =}\int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ A(x) \ dx$

$V = \int ^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ \pi (4- sec^2 (x) ) \ dx$

$V = 2 \pi \int ^{\dfrac{\pi}{3}}_{0}( 4 - sec^2 (x)) \ dx$

$V = 2 \pi \int ^{\pi/3}_{0} 4 . \ dx - 2 \pi \int ^{\pi/3}_{0} sec^2 (x) \ dx$

$V = 2 \pi(4) \int ^{\pi/3}_{0} 1 . \ dx - 2 \pi \Big( tan (x)\Big )^{\dfrac{\pi}{3}}_{0}$

$V = 8 \pi(x)^{\dfrac{\pi}{3}}_{0} - 2 \pi \Big( tan \dfrac{\pi}{3} -tan (0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}-{0}) - 2 \pi \Big( tan \sqrt{3}-(0)\Big )$

$V = 8 \pi({\dfrac{\pi}{3}}) - 2 \pi \Big( \sqrt{3}\Big )$

$\mathbf{V = 2 \pi \Big(\dfrac{4\pi}{3}- \sqrt{3} \Big)}$

8 0

2 years ago