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3 years ago

Help me pleasee this is to confusing.

Mathematics

1 answer:

AVprozaik [17]3 years ago

5 0

Answer:

What's confusing ??? You literally just put a dot on the number line

Example:

1.3, move 3 units to the right of 1 and place a dot there

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HELP PLEASE

GREYUIT [131]

The Answer is (D. No solution)

6 0

3 years ago

Twelve of the 40 cars pulled by the locomotive were tankers.

quester [9]

12 are tankesr
40 cars total
12/40
12/40=6/20=3/10

3/10=tankders
10/10-3/10=7/10
7/10=70/100=70%
70% wern't tankser

3 0

3 years ago

Check that 2b+b and 3b have the same value when b is 1,2, and 3. Do 2b+b and 3b have the same value for all values of b? Explain

Inessa [10]

Answer:

Expression 2b+b and 3b have same values for all values of b.

Step-by-step explanation:

Given the two expression $2b+b$ and $3b$ and values of b as 1, 2 and 3.

Consider first expression that is, $2b+b$ .

Substituting the value of $b=1$ ,

$2\left(1\right)+\left(1\right)=2+1=3$ ....1

Substituting the value of $b=2$ ,

$2\left(2\right)+\left(2\right)=4+2=6$ ....2

Substituting the value of $b=3$ ,

$2\left(3\right)+\left(3\right)=6+3=9$ ....3

Consider second expression that is, $3b$ .

Substituting the value of $b=1$ ,

$3\left(1\right)=3$ ....4

Substituting the value of $b=2$ ,

$3\left(2\right)=6$ ....5

Substituting the value of $b=3$ ,

$3\left(3\right)=9$ ....6

For value of b=1, equation 1 and equation 4 are same, for value of b=2, equation 2 and equation 5 are same and for value of b=3 equation 3 and equation 6 are same.

Therefore both expression are same for all values of b.

3 0

3 years ago

How do i find the area of this composite figure?

diamong [38]

To find the area of a composite figure, separate the figure into simpler shapes whose area can be found. Then add the areas together. Be sure than none of the simpler figures have overlapping areas

6 0

3 years ago

Read 2 more answers

How do you find the sine, cosine, and tangent values on the unit circle not in the first quadrant? Provide an example, and how y

gizmo_the_mogwai [7]

A point (-0.8, 0.6) will be a point on the unit circle in the second quadrant. Since it is a unit circle, its radius is 1, and we have
sin(α) = y = 0.6
cos(α) = x = -0.8
tan(α) = y/x = 0.6/-0.8 = -0.75

The angle is α = arccos(-0.8) ≈ 143.13°

______
For the unit circle, the trig values are always the coordinates or their ratio as shown above, regardless of quadrant.

7 0

3 years ago