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2 years ago

Directions: Complete a graphic organizer of twelve verbs which may either be regular or irregular.

Mathematics

1 answer:

Radda [10]2 years ago

5 0

Past form:

1. took

2. danced

3. taught

4. thought

5. twisted

6. kissed

7. shook

8. sought

9. touched

Past Participle:

1. taken

2. danced

3. taught

4. thought

5. twisted

6. kissed

7. shaken

8. sought

9. touched

You might be interested in

In the figure below, \overline{AD} AD start overline, A, D, end overline and \overline{BE} BE start overline, B, E, end overline

VLD [36.1K]

Answer:

The measure of the arc CD = 64°

Step-by-step explanation:

It is required to find the measure of the arc CD in degrees.

So, as shown at the graph

BE and AD are are diameters of circle P

And ∠APE is a right angle ⇒ ∠APE = 90°

So, BE⊥AD

And so, ∠BPE = 90° ⇒(1)

But it is given: ∠BPE = (33k-9)° ⇒(2)

From (1) and (2)

∴ 33k - 9 = 90

∴ 33k = 90 + 9 = 99

∴ k = 99/33 = 3

The measure of the arc CD = ∠CPD = 20k + 4

By substitution with k

∴ The measure of the arc CD = 20*3 + 4 = 60 + 4 = 64°

6 0

2 years ago

The height of a cylinder is twice the radius of its base. A cylinder has a height of 2 x and a radius of x. What expression repr

mrs_skeptik [129]

Given:

Let x represents the radius of the cylinder.

Given that the height of the cylinder is twice the radius of its base.

The height of the cylinder is 2x.

We need to determine the volume of the cylinder.

Volume of the cylinder:

The volume of the cylinder can be determined using the formula,

$V=\pi r^2h$

Substituting r = x and h = 2x, we get;

$V=\pi (x)^2(2x)$

Simplifying, we get;

$V=\pi x^2(2x)$

$V=2\pi x^3$

Thus, the expression that represents the volume of the cylinder is 2πx³ cubic units.

Hence, Option b is the correct answer.

8 0

2 years ago

Read 2 more answers

Assume that a fair die is rolled. The sample space is , 1, 2, 3, 4, 56, and all the outcomes are equally likely. Find P8. Expres

liraira [26]

Answer:

$P\left(8\right)=0$

Step-by-step explanation:

Assuming that a fair die is rolled.

The sample space is 1, 2, 3, 4, 5, 6 and all the outcomes are equally likely.

Let X be the set of all possible outcomes. Let A be an outcome.

So, the probability that A occurs is:

$P\left(A\right)=\frac{|A|}{|X|}$

As the set of all possible outcomes of the roll of a single die is:

$X=\left\{1,2,3,4,5,6\right\}$

Observe that

$|X|=6$

Here

$|A|=0$ because 8 is not in the set sample space. So, the outcome of occurring the number 8 is not possible from the all possible outcomes.

So, the probability must be zero.

In other words,

$P\left(A\right)=\frac{|A|}{|X|}$

$P\left(8\right)=\frac{0}{6}=0$

Therefore,

$P\left(8\right)=0$

5 0

2 years ago

Jacob spends 60 minutes in the gym every day doing freehand exercises and running on the treadmill. He runs on the treadmill 30

elixir [45]

Answer:

x + y = 60

x = y + 30

substitution:

(y + 30) = y = 60

2y + 30 = 60

2y = 30

y = 15

15 minutes for freehand

No, because if he spends 40 minutes on the treadmill, he would have spent 10 minutes for freehand exercises. His total time in the gym would be 50 minutes, not 60.

8 0

2 years ago

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 31% of courses have a res

beks73 [17]

Answer:

0.25

Step-by-step explanation:

72% of courses have final exams and 46% of courses require research papers which means probability of 0.72 for courses that have final exams and 0.46 for courses that require research papers.

31% of courses have a research paper and a final exam, which means probability of 0.31 for both courses with exams and research papers, using Venn diagram approach, find picture attached to the solution.

P(R or E) = P(R) + P(E) - P(R and E), which gives:

P(R or E) = 0.15 + 0.41 - 0.31

P(R or E) = 0.25.

6 0

2 years ago