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

PLS ANSWER CORRECTLY AND FAST I NEED TO SUBMIT PLS HELP PLS PLS PLS PLS......

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]

Quadrilaterals are plane shapes that are bounded by four straight sides. Thus, the required answers to the questions are:

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

47. False.

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

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

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

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

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