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2 years ago

20points please come help me

Mathematics

1 answer:

Snezhnost [94]2 years ago

4 0

Assuming these lines are perpendicular
Opposite angles are congruent therefore
77 = 6x + 23
6x = 54
X = 9

And supplementary angles sum up to be 180 degrees where in this case we can take angle 77 degrees and z which makes

77 + z = 180
Z = 103 degrees

Therefore,

x = 9 degrees
z = 103 degrees

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Consider the figure below. Abraham cut regular pentagon ABCDE into quadrilateral BEGC. Which of the following statements is true

baherus [9]

Answer:

Correct: 2, 3, 4, 6

Step-by-step explanation:

ABCDE is a regular pentagon. The sum of the measures of all interior angles in the pentagon is

$(n-2)\cdot 180^{\circ}=(5-2)\cdot 180^{\circ}=540^{\circ},$

then the measure of each interior angles in the regular pentegon is

$\dfrac{540^{\circ}}{5}=108^{\circ}$

Triangle ABe is isosceles triangle, so the adjacent to the bas BE angle has the measure

$m\angle ABE=\dfrac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$

Therefore,

$m\angle CBE=m\angle ABC-m\angle ABE=108^{\circ}-36^{\circ}=72^{\circ}$

(option 2 is correct)

Since $m\angle BCG=108^{\circ},$ then

$m\angle EGC=108^{\circ }+4x$

The sum of the measures of all interior angles in quadrilateral BCGE is

$(n-2)\cdot 180^{\circ}=(4-2)\cdot 180^{\circ}=360^{\circ},$

(option 6 is correct), then

$m\angle CBE+m\angle BEG+m\angle EGC+m\angle BCG=360^{\circ}\\ \\72^{\circ}+19x+3^{\circ}+108^{\circ}+4x+108^{\circ}=360^{\circ}\\ \\23x=69^{\circ}\\ \\x=3$

(option 1 is false)

$m\angle EGC=108^{\circ}+4\cdot 3^{\circ}=120^{\circ}$

(option 3 is correct)

$m\angle GCB+m\angle EGC=108^{\circ}+72^{\circ}=180^{\circ}$

(option 4 is correct)

$m\angle BEC+m\angle EGC=(19\cdot 3+3+108+4\cdot 3)^{\circ}=180^{\circ}$

(option 5 is false)

5 0

3 years ago

If k = (n + 2)(n – 2), where n is an integer greater than 2, what is the value of k ? (1) k is the product of two primes. (2) k

Korolek [52]

Answer:

The answer is k=77

Step-by-step explanation:

It is given that n is an integer and greater than 2. Let us simplify k = (n + 2)(n – 2) to k=n^2-4.

In (1) it is stated that k is the product of two primes and in (2) it is stated that k<100. To solve the question we need to consider both cases (1) and (2).

Let say n=9 then k=n^2-4 becomes k=81-4=77. Well, 77 is the product of two prime numbers 7 and 11 (77=7*11). The answer is k=77

6 0

3 years ago

P(5,-3) Q(2,4) find distance

Alenkinab [10]

Answer: $\sqrt{58}$