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Nikolay [14]
3 years ago
14

If all 6 Corresponding parts of 2 Triangles are congruent then both triangles are congruent

Mathematics
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

yes

Step-by-step explanation:


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astra-53 [7]

Step-by-step explanation:

  • 1

vector AB(3-(-6); 5-7)

vector AB(9;-2)

AB= \sqrt{9^{2}+(-2)^{2}  } = \sqrt{85}

  • 2

M is the midpoint of AB

we have B(-5;10) and M(1;7)

let A(x;y)

(x-5)/2 = 1 ⇒ x-5 = 2⇒ x = 7

(10=y)/2 = 7⇒ 10+y = 14 ⇒y= 4

so : A(7;4)

  • 3

the center of the circle is the midponit of the line joining both ends of the diameter

let A(x;y) be the other end

(-2+x)/2 = 2 ⇒ -2+x = 4⇒ x= 6

(5=y) = -1 ⇒ 5+y = -2 ⇒ y= -7

so the coordinates of the other end are (6; -7)

  • 4

A,B and C are collinear such as AB=BC so b is the midpoint of AC

(-5+1)/2 = y ⇒ y = -4/2 ⇒ y = -2

((-3=x)/2 = 7 ⇒ -3+x = 14 ⇒ x = 17

so x= 17 and y = -2

7 0
3 years ago
Help its due today ill do anythinggg
e-lub [12.9K]

Answer:

80.84

Step-by-step explanation:

3 0
3 years ago
Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.
mart [117]

Answer:

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

Step-by-step explanation:

The given expression is  

\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}    [\because a^ma^n=a^{m+n}]

\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}

\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}         [\because a^{-n}=\dfrac{1}{a^n}]

\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}.

5 0
3 years ago
I, so confused please help me asap!
Tamiku [17]

<u>Quadrilaterals</u> are <em>plane shapes</em> that are <u>bounded</u> by four <u>straight</u> sides. Thus, the required answers to the questions are:

46. True. Other examples include kites, rhombus, etc.

47. False.

46. When a <em>plane shape</em> is <u>bounded</u> by four <u>straight</u> sides of equal or different lengths, it is called a  <u>quadrilateral</u>. Examples include trapezium, kite, rhombus, rectangle, square, etc. Each of these examples has individual <u>properties</u>.

Thus the required answer to question 46 is; <u>True</u>. It can be observed that with respect to their <em>individual properties</em>, other <u>quadrilaterals</u> which have a pair of <em>opposite angles</em> to be <u>equal</u> include: kite, rectangle, rhombus, etc.

47. A <em>ray segment</em> is a given <u>line</u> that <u>points</u> or <u>heads</u> in a specific direction. So that the direction in which the ray moves is very<em> important</em>.

Thus in the given question, the <u>required</u> answer is; False. This is because the<u> two</u> given rays are moving in opposite directions. Though the two rays may have the <u>same</u> length of the <u>segment</u>.

For more clarifications on the properties of quadrilaterals, visit: brainly.com/question/21774206

#SPJ 1

7 0
2 years ago
In ΔCDE, \text{m}\angle C = (4x-16)^{\circ}m∠C=(4x−16) ∘ , \text{m}\angle D = (6x-1)^{\circ}m∠D=(6x−1) ∘ , and \text{m}\angle E
borishaifa [10]

Answer:

m∠C = 44°

Step-by-step explanation:

In ΔCDE,

m∠C=(4x−16) ∘

m∠D=(6x−1) ∘

m∠E=(4x−13) ∘ .

The sum of angles in a triangle = 180°

Step 1

We solve for x

m∠C + m∠D + m∠E

(4x−16)° + (6x−1)° + (4x−13)° = 180°

4x - 16 + 6x - 1 + 4x - 13 = 180°

4x + 6x + 4x -16 - 1 - 13 = 180°

14x - 30 = 180°

14x = 180+ 30

14x = 210

x = 210/14

x = 15

Step 2

Find m∠C

m∠C = (4x−16)°

m∠C = (4 × 15 - 16)°

m∠C = (60 - 16)°

m∠C = 44°

3 0
3 years ago
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