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3 years ago

Use this triangle.

1 answer:

5 0

The sides are in the ratio 1 : 1 : sqrt2 where sqrt 2 = hypotenuse

so if the length of one of the legs is 2 cm the hypotenuse will be of length 2 *sqrt 2 cms.

b ) sin 45 = 1 / sqrt2 = sqrt2 / 2

c) if the 2 legs of triangle are both 2 cms area will be 1/2* 2* 2

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Find the length of x

zaharov [31]

Answer:

The answer is 2

Step-by-step explanation:

The hypotenuses of the triangles are different. One is 4 and the other is 5.

Lets use ratios to find the relationship of the hypotenuse and the base.

2.5:5

Turn this into a fraction

2.5/5

Now divide

2.5/5 = 0.5

The base of the triangle is half the length of the hypotenuse.

So 4 is x’s hypotenuse so:

4 x 0.5 = 2

X = 2

6 0

2 years ago

Can someone translate this into a verbal expression 3(4j + 4 + j)?

dalvyx [7]

3(5j+4)
15j+12

final answer 15j+12

5 0

3 years ago

Multiple<br>1. Which of the following is Eular's formula?<br>OE-V + F = 2<br>OV-E+ F = 2

Alinara [238K]

Answer:

see explanation

Step-by-step explanation:

Euler's formula for polyhedra is

V- E + F = 2

where V is number of vertices, E number of edges and F number of faces

6 0

3 years ago

What is the solution to an equation /

navik [9.2K]

Answer:

Idk what u mean?

Step-by-step explanation:

but i searched it up and here are my results

The solution of an equation is the set of all values that, when substituted for unknowns, make an equation true. For equations having one unknown, raised to a single power, two fundamental rules of algebra, including the additive property and the multiplicative property, are used to determine its solutions.Mar 20, 2020

email me if it helps

6 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