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3 years ago

9

You have been cleaning diesel dispensers for 35 minutes. You have completed 6 dispensers. How many minutes has it taken you to c

lean 1 dispenser?

2 answers:

Yanka [14]3 years ago

4 0

Just divide the number of minutes it took to clean 6 dispensers by 6.

35 / 6 = 5.83

So it would take about 5.83 minutes to clean one dispenser.

balu736 [363]3 years ago

3 0

It takes 5.8 minutes. Divide the time, 35 minutes by the amount of dispensers, 6.

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Error Analysis jessie incorrectly said the rate 1/5 hour 1/10 acer can be written as the unit rate 1/50 per acer . What is the c

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2 years ago

The following question has two parts. First, answer part A. Then, answer part B.

Pie

Answer:

When you're talking factors, you're talking about some sort of integer; that's because “factors” depends on the concept of divisibility, which are virtually exclusive to integers. When you're talking “greater than”, you're excluding complex numbers (where the concept of ordering doesn't exist) and you're probably assuming positive integers. If you are, then no; no positive integer has factors that are larger than it.

If you go beyond positive numbers, that changes. 0 is an integer, and has every integer, except itself, as factors; since its positive factors are greater than zero, there are factors of zero that are greater than zero. If you extend to include negative numbers, you always have both positive and negative factors; and since all positive integers are greater than all negative integers, all negative integers have factors that are greater than them.

Beyond zero, though, no integer has factors whose magnitudes are greater than its own. And that's a principle that can be extended even to the complex integers

Step-by-step explanation:

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2 years ago

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HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME

Kryger [21]

Answer:

1)

Given the triangle RST with Coordinates R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.

Since, the scale factor $\frac{5}{3}$ is greater than 1, the image is enlargement or a stretch.

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor $\frac{5}{3}$

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image $S{}'$ are same.

Now, the distances from the center of the dilation at point S to the other points R and T.

The dilation image will be $\frac{5}{3}$ of each of these distances,

$SR=3$ , so $S{}'R{}'$ =5 ;

$ST=3$ , so $S{}'T{}'=5$

Now, draw the image of RST i.e R'S'T'

Since, $RT=3\sqrt{2}$ [By using hypotenuse of right angle triangle] and $R{}'T{}'=5\sqrt{2}$ .

2)

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of $\overline{AC}$ i.e $\overline{AB}\cong \overline{BC}$

In the triangle ABD and triangle CBD, we have

$\overline{AB}\cong \overline{BC}$ (SIDE) [Given]

$\overline{BD}\cong \overline{BD}$ (SIDE) [Reflexive post]

Since, there is no included angle in these triangles.

∴ $\Delta ABD$ is not congruent to $\Delta CBD$ .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that $\overline{AD}\cong \overline{CD}$ .

therefore, in the triangle ABD and triangle CBD, we have

$\overline{AB}\cong \overline{BC}$ (SIDE) [Given]

$\overline{AD}\cong \overline{CD}$ (SIDE) [Given]

$\overline{BD}\cong \overline{BD}$ (SIDE) [Reflexive post]

therefore by, SSS postulates $\Delta ABD\cong \Delta CBD$ .

3)

Given that: $\angle1=\angle 3$ are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

$\angle 1$ and $\angle 2$ and $\angle 3$ and $\angle 2$ are linear pair.

By linear pair theorem, $\angle 1$ and $\angle 2$ are supplementary, $\angle 2$ and $\angle 3$ are supplementary.

$m\angle1+m\angle 2=180^{\circ}$

$m\angle2+m\angle 3=180^{\circ}$

Equate the above expressions:

$m\angle 1+m\angle 2=m\angle 2+m\angle 3$

Subtract the angle 2 from both sides in the above expressions

∴ $m\angle 1=m\angle 3$

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.

therefore, $\angle 1\cong \angle 3$ .

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2 years ago

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