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3 years ago

Which fraction is bigger 2/6 or 2/4

1 answer:

3 0

2/4 is the larger fraction. Think of a candy bar that is cut into 4 equal pieces. Now think of the same sized candy bar cut into 6 pieces. The candy bar that has been cut into sixths has more pieces but they are smaller pieces that the candy bar cut into fourths. **When comparing fractions, if the numerators are the same (like they are in this problem) look at the denominators. Whichever fraction has the greater number is actually going to be the smaller fraction (more pieces, but smaller pieces). If the denominators are the same, then the fraction that has the greater number in the numerator will be the larger fraction (more same-sized pieces than the other fraction) Hope that helps!

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Help with this question please

Maru [420]

Answer:

I don't know so sorry!

5 0

3 years ago

Question and choices are in the photo please explain the answer

Natasha_Volkova [10]

Answer:

$\text{A) }68\:\mathrm{cm}$

Step-by-step explanation:

The perimeter of a polygon is equal to the sum of all the sides of the polygon. Quadrilateral PTOS consists of sides TP, SP, TO, and SO.

Since TO and SO are both radii of the circle, they must be equal. Thus, since TO is given as 10 cm, SO will also be 10 cm.

To find TP and SP, we can use the Pythagorean Theorem. Since they are tangents, they intersect the circle at a $90^{\circ}$ , creating right triangles $\triangle TOP$ and $\triangle SOP$ .

The Pythagorean Theorem states that the following is true for any right triangle:

$a^2+b^2=c^2$ , where $c$ is the hypotenuse, or the longest side, of the triangle

Thus, we have:

$10^2+TP^2=26^2,\\TP^2=26^2-10^2,\\TP^2=\sqrt{576},\\TP=24$

Since both TP and SP are tangents of the circle and extend to the same point P, they will be equal.

What we know:

$TP=SP=24$
$TO=SO=10$

Thus, the perimeter of the quadrilateral PTOS is equal to $24+24+10+10=\boxed{\text{A) }68\:\mathrm{cm}}$

7 0

3 years ago

Read 2 more answers

Evaluate (f + g)(x) if f(x) = 2x and g(x) = 3X - 2<br> when x = 3

77julia77 [94]

Answer:

Step-by-step explanation:

→ Substitute 3 into 2x

2 × 3

→ Evaluate

f ( x ) = 6

→ Substitute x = 3 into 3x - 2

3 × 3 - 2

→ Evaluate

→ Find the sum of the 2 results

7 0

2 years ago

(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)

grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0

2 years ago

Please help! Which expression is equal to..?

Alenkasestr [34]

Answer:

Step-by-step explanation:

Before we can add the 2 fractions we require them to have a common denominator.

Multiply the numerator/ denominator of the first fraction by x

Multiply the numerator/ denominator of the second fraction by (x + 3)

= $\frac{3x(x)}{x(x+3)}$ + $\frac{(x-2)(x+3)}{x(x+3)}$

Now simplify the numerator, leaving the denominator

= $\frac{3x^2+x^2+x-6}{x(x+3)}$

= $\frac{4x^2+x-6}{x(x+3)}$ → A

5 0

3 years ago