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3 years ago

Which congruency theorem can be used to prove that ABD =DCA?

Mathematics

1 answer:

Dimas [21]3 years ago

6 0

Answer:

The answer is "SAS".

Step-by-step explanation:

In this question we assume two angle is given that are:

$\ Two \ triangles \angle ABD \ and \angleDCA \\\\\ In \ this \ on \ angle \ is \ common \ that \ is, \\\\ AD=AD \\\\AB= CD \\\\\angle A \ is \ equal \ = \angle A \ (\ which \ is \ given)\\\\$

That's why SAS rule is correct.

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Ocean Bike Tours wants to provide bandanas for each person. The cost of the bandanas is each person. The cost of the bandana is

mixas84 [53]

Answer:

$145.50

Step-by-step explanation:

C = 45.50 + 2n

45.50 + 2X 100 = $145.50

7 0

3 years ago

Factorise x² - x with explanation please?

Assoli18 [71]

Answer:

x(x + 1)

Step-by-step explanation: This is an important way of solving quadratic equations. The first step of factorising an expression is to 'take out' any common factors which the terms have. So if you were asked to factorise x² + x, since x goes into both terms, you would write x(x + 1)

3 0

2 years ago

The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?

Jet001 [13]

Answer:

The roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question, we have that:

$2x^2 + x - 6 = 0$

Which is a quadratic equation with $a = 2, b = 1, c = -6$ . So

$\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49$

$x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}$

$x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2$

So the roots are $x = (\frac{3}{2}, -2)$ , which is given by option C.

3 0

3 years ago

An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had

Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2 CI 95 % = ( 0,056 ; 0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁ = 500

x₁ = 180

p₁ = 180/ 500 p₁ = 0,36 then q₁ = 1 - p₁ q₁ = 0,64

Sample information for people over 18

n₂ = 600

x₂ = 150

p₂ = 150 / 600 p₂ = 0,25 then q₂ = 1 - p₂ q₂ = 1 - 0,25 q₂ = 0,75

Hypothesis Test

Null hypothesis H₀ p₁ = p₂

Alternative Hypothesis Hₐ p₁ ≠ p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI = 95 % significance level is α = 5 % α = 0,05 and

α/ 2 = 0,025 z (c) for that α is from z-table:

z(c) = 1,96

To calculate z(s)

z(s) = ( p₁ - p₂ ) / EED

EED = √(p₁*q₁)n₁ + (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500 + (0,25*0,75)/600

EED = √0,00046 + 0,0003125

EED = 0,028

( p₁ - p₂ ) = 0,36 - 0,25 = 0,11

Then

z(s) = 0,11 / 0,028

z(s) = 3,93

Comparing z(s) and z (c) z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2 CI 95 % = ( p₁ - p₂ ) ± z(c) * EED

CI 95% = ( 0,11 ± 1,96 * 0,028 )

CI 95% = ( 0,11 ± 0,054 )

CI 95 % = ( 0,056 ; 0,164 )

8 0

2 years ago

Of th 30 girls who tried out for the lacrosse team team at Euclid middle school , 12 were selected. Of the 40 boys show tried ou

Harman [31]

12/30 is simplified to 2/5 and<span>number of students trying out the same for both boys and girls</span> 16/40 is also simplified to 2/5 so YES the number of students trying out is the same for both boys and girls. I hope this helps :)

3 0

3 years ago