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Rasek [7]
3 years ago
8

The first step in solving 3.17 t - 2.431 = 5.06 is to divide both sides of the equation by 3.17. TrueFalse

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
5 0

Answer:

False

Step-by-step explanation:

3.17t - 2.431 = 5.06 \\ 3.17t = 5.06 + 2.431 \\ 3.17t = 7.491 \\  \frac{3.17t}{3.17}  =  \frac{7.491}{3.17}  \\ t = 2.363

andrew-mc [135]3 years ago
3 0

Answer:

False Just False NOT TRUE just False

Step-by-step explanation:

You might be interested in
Find mRQT.<br> (4x + 15)<br> (10x - 3)
ELEN [110]

Answer:

61 degrees

Step-by-step explanation:

The angle measure must be less than 90 degrees (4x+15) (10x-3) is 61 degrees

3 0
3 years ago
How can you predict whether the sum of two integers is 0, positive, or negative?
zepelin [54]

Answer:

See explanation:

Step-by-step explanation:

Let a and b be integers.

Sum means addition.

So we are trying to figure out what a+b equals.

It could result in as 0, negative, or positive.

a+b is 0 when a and b are of opposite values. Example: 5+(-5) or -5+5 is 0 because 5 and -5 are opposite values.

a+b is positive when both a and b are positive. Example 5+3=5.

a+b is positive when |a|>|b| and a is positive. Example:  5+(-3)=2 since |5|>|-3| and 5 is positive.

a+b is positive when |a|<|b| and b is positive. Example:  -3+5=2 since |-3|<|5| and 5 is positive.

a+b is negative when both a and b are negative. Example: -5+(-3)=-8.

a+b is negative when |a|>|b| and a is negative. Example: -5+3=-2 since |-5|>|3| and -5 is negative.

a+b is negative when |a|<|b| and b is negative. Example: 3+(-5)=-2 since      |3|<|-5| and -5 is negative.

8 0
3 years ago
Please help thank you.
Wittaler [7]
30 60 90 right triangle
leg = 6 so other leg = 6√3

A = (1/2)(6+6)( 6√3+ 6√3)
A = 6 (12√3)
A = 72√3

answer is D
72√3

7 0
3 years ago
I need help finding this answer please
Maslowich
To solve this you need to use the formula for Chord-Chord length. You need to know it, it is ab=cd. with this, we can plug numbers in.
a=30
b=x+1
c=2x-6
d=21
So our new equation is:
30(x+1)=21(2x-6)
30x+30=42x-126
156=12x
x=13
This isn't what the question is askign for though, it wants YZ
Which is A+B or 30+x+1 or 21+13=34. 
YZ=34

8 0
3 years ago
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
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