Answer:
1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻
Step-by-step explanation:
1. Let b be the number and B be the set, so mathematically, it is written as
b ∈ B.
2. Let a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write
∀ a ∈ N; 2a ∈ Z
3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.
Since each set is a subset of the latter set, we write
N ⊂ Z ⊂ Q ⊂ R .
4. Let J be the negative integer which is an element if negative integers. Let the set of negative integers be represented by Z⁻. Since J is less than or equal to its inverse, we write
J ≤ J⁻¹ : J ∈ Z⁻
Step-by-step explanation:
SAS is Side - Angle (between the sides) - Side
BC and CD are shown to be congruent with the markers that cross the line segments.
AC and AC are the same because they are the same.
The only things needed are the congruent angles between the sides.
The angles are ACD and ACB, because those are the angles between AC and BC and AC and CD.
Answer:
The answers to each part are:
Part A.
- <u>The quantity of apples is one-third of the quantity of grapes</u>.
Part B.
- <u>The quantity of apples is a quarter of the quantity of strawberries</u>.
Part C.
- <u>The number of cherries is two-elevenths of the total fruit</u>.
Step-by-step explanation:
To identify the answer in each case, you must remember that all the parts are equal, then:
Part A.
The parts of apples are 1 and the parts grapes are 3, so if you divide the first quantity with the second quantity you obtain:
So, <u>the quantity of apples is one-third of the quantity of grapes</u> or the quantity of apples is three times smaller than the quantity of grapes.
Part B.
The parts of apples are 1 and the parts of strawberries are 4, then you must divide the first quantity with the second quantity:
In this case, <u>the quantity of apples is a quarter of the quantity of strawberries</u> or the quantity of apples is four times smaller than the quantity of strawberries.
Part C.
First, you must add all the part of fruit:
- <em>1 part apple</em>
- <em>1 part orange</em>
- <em>4 parts strawberry</em>
- <em>2 parts cherry </em>
- <em>3 parts grape</em>
The total of fruits is 11 parts, taking into account the quantity of cherries is 2, now you can divide the number of cherries with the total parts of fruit:
- 2 / 11 = 2/11 (two-elevenths)
Now, you can see <u>the number of cherries is two-elevenths of the total fruit</u>.
Answer:
y=|x|
Here's a graph if needed.
→ 6x² - 2x - 20
→ 6x² + 10x - 12x - 20
→ x(6x + 10 ) - 2 (6x + 10)
→ (6x + 10) (x - 2) .
for
6x - 10 = 0
6x = 10
for x - 2
x = 2
