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3 years ago

Can any one slove this

Mathematics

1 answer:

Rashid [163]3 years ago

3 0

The exterior angle is = to the 2 interior angles.
83 = (9x + 4) + (4x + 1) Remove the brackets
83 = 9x + 4 + 4x + 1 Group the like terms
83 = 9x + 4x + 4 + 1 Add the like terms.
83 = 13x + 5 Subtract 5 from both sides.
78 = 13x Divide by 13
78 / 13 = x
x = 6

So They want the angles
9x + 4 = 9*6 + 4 = 58
4x + 1 = 4*6 + 1 = 25

When you add these two together you should get 83. Do you?

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Compute and compare the side lengths of the two figures. Does this support an argument claiming that they are congruent? What el

Sunny_sXe [5.5K]

We compute for the side lengths using the distance formula √[(x₂-x₁)²+(y₂-y₁)²].

AB = √[(-7--5)²+(4-7)²] = √13
A'B' = √[(-9--7)²+(0-3)²] = √13

BC = √[(-5--3)²+(7-4)²] = √13
B'C' = √[(-7--5)²+(3-0)²] =√13

CD = √[(-3--5)²+(4-1)²] = √13
C'D' = √[(-5--7)²+(0--3)²] = √13

DA = √[(-5--7)²+(1-4)²] = √13
D'A' = √[(-7--9)²+(-3-0)²] = √13

The two polygons are squares with the same side lengths.

But this is not enough information to support the argument that the two figures are congruent. In order for the two to be congruent, they must satisfy all conditions:
1. They have the same number of sides.
2. All the corresponding sides have equal length.
3. All the corresponding interior angles have the same measurements.

The third condition was not proven.

6 0

3 years ago

Read 2 more answers

Factor completely: 64 - y^3

madam [21]

From your equation, you can see that you have a difference of two cubes (aka two cubes being subtracted): 64, which is $4^{3}$ , and $y^{3}$ .

There is rule for the difference of two cubes:
The difference of two cubes is equal to the difference of the cube roots times a binomial, which is the sum of the squares of the roots plus the product of the roots.

That sounds pretty confusing, but it's much easier to understand when put mathematically. Let's say our two cubes are $a^{3}$ and $b^{3}$ . The difference of those two cubes is:
$a^{3} - b^{3} = (a - b)( a^{2} + ab + b^{2})$

In our problem, a = 4 (since $a^{3}$ = 64) and b = y (since $b^{3} = y^{3}$ . Plug these values into the rule to find the factor of $64 - y^3$ :
$64 - y^3 \\
= (4 - y)( 4^{2} + 4y + y^{2}) \\
= (4 - y)( 16 + 4y + y^{2})$

-----

Answer: $(4 - y)( 16 + 4y + y^{2})$

8 0

3 years ago

Plz help me 56+(96x3)-34=?

Veseljchak [2.6K]

The answer to it is 310

7 0

3 years ago

Read 2 more answers

Classify the following triangle. Check all that apply

lubasha [3.4K]

Answer:

A. Acute and B. Equilateral

Step-by-step explanation:

8 0

2 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

2 years ago