HELP ME PLEASE!

1 answer:

7 0

Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.

Let's see whether they do.

8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.

If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:

8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and

d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.

11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.

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3 years ago

[need math help] Study buddies part #2

rosijanka [135]

Answer:

Part A) The distance between Whitney's house and Anh's house is 6.5 miles.

Part B) The total distance traveled by Whitney during the trip was 9.5 miles.

Step-by-step explanation:

Part A)

Notice that during Whitney's journey to Anh's house, her velocity was negative for some time. So, Whitney was travelling backwards. Hence, the total distance Whitney traveled will be greater than the actual distance from Whitney's house to Anh's.

So, instead, we can calculate Whitney's total displacement. Total displacement is given by:

$\displaystyle \text{Displacement}=\int_a^bv(t)\, dt$

Where v(t) is the velocity function.

So, the displacement of Whitney when she traveled from her house to Anh's house will be:

$\displaystyle \text{Displacement}=\int_0^{24}v(t)\, dt$

To calculate the integral, we can split it off at t = 11, t = 16, and t = 24.

The three regions represent three geometric figures, which we can use to calculate the area and then find the integral. Rewriting:

$\displaystyle \text{Displacement}=\int_0^{11}v(t)\, dt+\int_{11}^{16}v(t)\, dt+\int_{16}^{24}v(t)\, dt$

Now, we can calculate the area of each figure.

The first figure is a trapezoid with a height of 0.8 and two bases of 11 and 3. So, its area is:

$\displaystyle A_1=\frac{1}{2}(0.8)(11+3)=5.6\text{ miles}$

The second figure is a triangle with a height of 0.6 and a base of 5. So, its area is:

$\displaystyle A_2=\frac{1}{2}(0.6)(5)=1.5\text{ miles}$

Lastly, the third figure is another triangle will a height of 0.6 and a base of 8. So, its area is:

$\displaystyle A_3=\frac{1}{2}(0.6)(8)=2.4\text{ miles}$

However, notice that the second figure is below the x-axis. During that interval, Whitney was traveling backwards. Hence, we will subtract the second area from the rest of the areas.

Then the displacement will be:

$\displaystyle \text{Displacement}=5.6-1.5+2.4=6.5\text{ miles}$