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3 years ago

Decompose fraction 11/5

1 answer:

6 0

Hi there, bascially decompose means two fractions that add up to 11/5. There are alot of ways to decompose 11/5 so I am only giving 3 ways to decompose 11/5. The first way is 10/5+1/5, the second way is 9/5+2/5, and the third way is 6/5+5/5

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Can someone help me with this please

Bingel [31]

Answer:

A.1: ∠BAC ≅ ∠BDC ≅ ∠EDF, ∠ACD ≅ ∠ABD ≅ ∠BDE ≅ ∠CDF

A.2: ∠1 ≅ ∠4, ∠2 ≅ ∠3 ≅ ∠5 ≅ ∠6

A.3: ∠2 ≅ ∠3

B.1: ∠ACD ≅ ∠CAB, ∠CDA ≅ ∠ABC, ∠DAC ≅ ∠BCA

B.2: ∠1 ≅ ∠3 ≅ ∠5, ∠2 ≅ ∠4 ≅ ∠6

see "additional comment" regarding listing pairs

Step-by-step explanation:

There are a number of ways angles can be identified as congruent. In each case, the converse of the proposition is also true.

opposite angles of a parallelogram are congruent
corresponding angles where a transversal crosses parallel lines are congruent
alternate interior angles where a transversal crosses parallel lines are congruent
vertical angles are congruent
any two angles with the same measure are congruent

In these exercises, pairs of angles need to be examined to see which of these relations may apply.

Left

ABCD is a parallelogram, so the congruent angles are opposite angles and any that are vertical or corresponding:

∠BAC ≅ ∠BDC ≅ ∠EDF ≅ 110° (3 pairs)

∠ACD ≅ ∠ABD ≅ ∠BDE ≅ ∠CDF ≅ 70° (6 pairs)

Center

∠1 ≅ ∠4 ≅ 66° (1 pair) . . . . vertical angles

∠2 ≅ ∠3 ≅ ∠5 ≅ ∠6 ≅ 57° (6 pairs) . . . . marked with the same measure, and their vertical angles

Right

Assuming that lines appearing to go in the same direction actually do go in the same direction, the only pair of congruent angles in the figure is ...

∠2 ≅ ∠3

Left

Corresponding angles in congruent triangles are congruent. Here, the congruent triangles are ΔACD ≅ ΔCAB. So, the pairs of congruent angles are ...

∠ACD ≅ ∠CAB (30°)

∠CDA ≅ ∠ABC (90°)

∠DAC ≅ ∠BCA (60°)

Right

The corresponding angles and any vertical angles are congruent. This means all the odd-numbered angles in the figure are congruent, and all the even-numbered angles in the figure are congruent. The marked 72° angles show the "horizontal" segments are parallel by the converse of the corresponding angles theorem.

∠1 ≅ ∠3 ≅ ∠5 (72°) (3 pairs)

∠2 ≅ ∠4 ≅ ∠6 (108°) (3 pairs)

_____

Additional comment

The question asks you to list pairs of congruent angles. When 3 things are congruent, they can be arranged in 3 pairs:

a ≅ b ≅ c ⇒ (a≅b), (a≅c), (b≅c)

Similarly, when 4 things are congruent, they can be arranged in 6 pairs:

a ≅ b ≅ c ≅ d ⇒ (a≅b), (a≅c), (a≅d), (b≅c), (b≅d), (c≅d)

In the above, we have elected not to list all of the pairs, but to list the set of congruences from which pairs can be chosen.

6 0

3 years ago

PLS HELP ME IMMA CRY ONGG GGHGHJH

WITCHER [35]

Answer:

127

Step-by-step explanation:

We started at 7

We keep going up 3

multiply

$3\cdot40=120$ Now add $120+7=127$ at the 40th term

8 0

3 years ago

Read 2 more answers

How many real sixth roots are in 15,625?

Murljashka [212]

There are two real roots (5 and -5) and 4 imaginary roots

3 0

3 years ago

11. Tickets to a football game were $5 for

slamgirl [31]

Answer:

The number of children that attended the game was 365

Step-by-step explanation:

Let

x ----> the number of children that attended the game

y ---> the number of adults that attended the game

we know that

$x+y=587$ ----> equation A

$5x+7y=3,379$ ----> equation B

Solve the system by graphing

Remember that the solution of the system is the intersection point both graphs

The solution is the point (365,222)

see the attached figure

therefore

The number of children that attended the game was 365

4 0

3 years ago

the vertex of this parabola is at (2,-4). when the y-value us -3, the x-value is -3. what is the coefficient of the squared term

alex41 [277]

Answer:

The coefficient of the squared term is 1/25.

Step-by-step explanation:

We are given that the vertex of a parabola is at (2, -4). We also know that y = -3 when x = -3.

And we want to determine the coefficient of the squared term of the equation.

Since we are given the vertex, we can use the vertex form of the quadratic:

$\displaystyle y = a(x-h)^2+k$

Where (h, k) is the vertex and a is the leading coefficient. The leading coefficient is also the coefficient of the squared term, so we simply need to find the value of a.

Since the vertex is at (2, -4), h = 2 and k = -4. Substitute:

$\displaystyle y = a(x-2)^2-4$