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3 years ago

An angle measures 46° more than the measure of its supplementary angle. What is the measure of each angle?

Mathematics

1 answer:

Alona [7]3 years ago

6 0

Answer:

67 degree

113 degree

Step-by-step explanation:

Supplementary angles are angles that measure up to 180 degrees. For example, if one angle measures 60 degree, the other angle would measure (180 - 60) 120 degrees

Let x represent the smaller angle

The bigger angle = x + 46

We know that both the smaller and the bigger angle would add up to 180, thus this equation can be written :

x + 46 + x = 180

2x + 46 = 180

Collect like terms

2x = 180 - 46

2x = 134

Divide both sides by 2

x = 67 degrees

the bigger angle = 67 + 46 = 113 degree

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What must be true of the units in two rates if one rate can be converted to the units of the other rate? In other words, the uni

puteri [66]

Answer:

We have two rates, A and B.

If we want that the units of A can be converted into the units of B, then we must have that the units of A represent the same measure that the units of B.

What does this mean?

For example, if A is written in distance over time, then B must also must have units of distance over time.

An example would be:

units for distance = meter, kilometer, feet, etc.

units for time = hour, minute, second, etc-

Then we could have:

[A] = [km/h]

[B] = [m/min]

Then the units of A could be converted into the ones of B, because both of them have the same physical meaning.

Concluding, you only can convert units that represent the same physical quantity, for example, if you have one quantity in meters, for example:

10m

You only can convert this to other units of distance, remember to do the correct conversion, if we would want to write this in cm, you should remember that:

1m = 100cm

Then, 10 times one meter is 10 times 100 cm

10m = 10*(1m) = 10*(100cm) = 1000cm

7 0

3 years ago

1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given

neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

$x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}$

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: $x^2-2x-4=0$

$x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}$

As for

$x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\$

3) Rewriting and calculating its derivative. Remember to do it, in radians.

$5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1$

$x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}$

For the second root, let's try -1.5

$x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\$

For x=-3.9, last root.

$x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\$

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

$x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}$

$f(x)=x^6-x^4+3x^3-2x$

$\mathbf{f'(x)=6x^5-4x^3+9x^2-2}$

$\mathbf{f''(x)=30x^4-12x^2+18x}$

For -1.2

$x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx \mathbf{-1.29322}\\$

For x=0.4

$x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx \mathbf{0.50785}\\$

and for x=-0.4

$x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\$

These roots (in bold) are the critical numbers

3 0

2 years ago

I need help with this! Quick ASAP?

sp2606 [1]

Answer:

a) F

b) B, E, D

Step-by-step explanation:

a) The segment with the greatest gradient has the largest change in y-values per unit change in x-values

From the given option, the rate of change of the <em>y </em>to the<em> </em>x-values of B = the gradient = (4 units)/(2 units) = 2

The gradient of F = (-3units)/(1 unit) = -3

The gradient of A = 4/4 = 1

The gradient of C = -2/5

The gradient of D = 2/6 = 1/3

The gradient of E = 3/4

The segment with the greatest gradient is F

b) The steepest segment has the higher gradient

From their calculated we have;

The gradient of segment B = 2 therefore, B is steeper than E that has a gradient of 3/4, and E is steeper than D, as the gradient of D = 1/3

Therefore, we have;

B, E, D.

6 0

3 years ago

What is in the middle of 0 and 1

Mashcka [7]

Answer:

Step-by-step explanation:

That 0 between 1 is .5

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Answer:

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4 0

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