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3 years ago

Use the Pythagorean theorem to find the length of the line

Mathematics

1 answer:

Elanso [62]3 years ago

6 0

Answer:

This is one of my specialties so sit back, relax while I work on this for 5 mins :)

Step-by-step explanation:

The formula is a^2 + B^2 = c^2

Sooo

7^2 + 5^2 = c^2

49 + 25 = 74

√74 = √c^2

8.60232526 = c

8.60232526 rounded to the nearest tenth is 8.6

So there you go. Hope it helps!!

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In this diagram, ABAC – AEDF. If the<br> area of ABAC = 6 in?, what is the<br> area of AEDF?

shepuryov [24]

Answer:

2.7 in²

Step-by-step explanation:

similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.

so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.

in other words

EF = BC × 2/3

2 = 3 × 2/3

correct

how do we calculate the area of a triangle ?

Area = baseline × height / 2

from BAC we know

Area = 6

baseline = 3

height = ?

6 = 3 × height / 2

12 = 3 × height

height = 4

aha !

now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3

so, for EDF we know

Area = ?

baseline = 2

height = 4 × 2/3 = 8/3

therefore, the area is

Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7

the shirt answer would be :

we know from the 2 baselines that the scaling factor for each distance is 2/3.

for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.

2/3 × 2/3 = 4/9

Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7

3 0

3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

3 years ago

In how many ways can 12 books be displayed on a shelf if the given number of books are available

Cerrena [4.2K]

The answer is not applicable. theres not enough information. -fvmousskaylaa

4 0

3 years ago

Ellie is renting a paddle boat. The boat rents for a 5 plus 3.50 per hour. She wants to spend under $30 altogether Write an ineq

Arada [10]

3.50x + 5 < 30

Ellie wants to spend less than 30, so it’ll be less than.

4 0

3 years ago

11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0

NISA [10]

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

Answer:

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

4 0

3 years ago