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3 years ago

10

The image shown has two triangles sharing a vertex:

2 answers:

Vika [28.1K]3 years ago

8 0

Answer:

40°.

Step-by-step explanation:

We can see from the diagram (in the comments) that ∠UWV and ∠XWZ are vertical angles; this is because they are opposite angles that share only a vertex.

We can also see that both of these angles are 90°.

This gives us two angle measures of triangle WXZ, 50 and 90. We know that the sum of the measures of the angles of a triangle is 180°; this gives us

180-(50+90) = 180-140 = 40° for the angle.

ZanzabumX [31]3 years ago

6 0

So you know that the sum of angles in a triangle is 180°.
You know two of the angles in triangle WXZ. The two angles are angle WXZ, which is 50° and angle XWZ, which is 90°.

Now you can calculate angle WZX, by:
180-50-90=40°.

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3 years ago

What is 3X^2 × 2X^3

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3 years ago

A rectangle is four times as long as it is wide. The perimeter is 40 cm. Find the length of each side.

rjkz [21]

Answer:

W=4 l=16

Step-by-step explanation:

Let w = width

Let l = length

Since the length is 4 times the width, l=4w

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= 2(4w)+2w

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6 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago