Ask question

Ask question

All categories

3 years ago

13

Given: AD ≅ BC and AD ∥ BC Prove: ABCD is a parallelogram. Statements Reasons 1. AD ≅ BC; AD ∥ BC 1. given 2. ∠CAD and ∠ACB are

alternate interior ∠s 2. definition of alternate interior angles 3. ∠CAD ≅ ∠ACB 3. alternate interior angles are congruent 4. AC ≅ AC 4. reflexive property 5. △CAD ≅ △ACB 5. SAS congruency theorem 6. AB ≅ CD 6. ? 7. ABCD is a parallelogram 7. parallelogram side theorem What is the missing reason in step 6?

2 answers:

tiny-mole [99]3 years ago

6 0

Given: It is given that $AD \cong BC$ and $AD \parallel BC$ .

To Prove: ABCD is a parallelogram.

Statements

1. $AD \cong BC$ and $AD \parallel BC$

Reason: Given

2. $\angle CAD, \angle ACB$ are alternate interior angles

Reason: Def of alternate interior angles

3. $\angle CAD \cong \angle ACB$

Reason: Alternate interior angles are congruent

4. $AC \cong AC$

Reason: Reflexive property

5. $\Delta CAD \cong \Delta ACB$

Reason: SAS congruency theorem

6. $AB \cong CD$

Reason: Corresponding Parts of Congruent triangles are Congruent (CPCTC)

7. ABCD is a parallelogram

Reason: Parallelogram Side Theorem

ira [324]3 years ago

5 0

Given: AD ≅ BC and AD ∥ BC

Prove: ABCD is a parallelogram.

Statements Reasons

1. AD ≅ BC; AD ∥ BC 1. given

2. ∠CAD and ∠ACB are alternate interior ∠s 2. definition of alternate interior angles

3. ∠CAD ≅ ∠ACB 3. alternate interior angles are congruent

4. AC ≅ AC 4. reflexive property

5. △CAD ≅ △ACB 5. SAS congruency theorem

6. AB ≅ CD 6. Corresponding Parts of Congruent triangles are Congruent (CPCTC)

7. ABCD is a parallelogram 7. parallelogram side theorem

<span> </span>

You might be interested in

Given a two-digit number, make a three digit number by putting a 6 as the rightmost digit. Then, add 6 to the resulting three di

VladimirAG [237]

By working backwards,
<span>you get 76 as your answer because you have to remove the rightmost digit, right? so the three-digit number is </span>

<span>76x </span>
<span>(x stands for the rightmost digit, because we don't know what the rightmost digit is) </span>

<span>we are told we have to add six to the three digit number to get 76x, so we have to subtract 6 from 76x: </span>
<span>76x - 6 = ??? </span>

<span>since we assume that the original two-digit number is NOT 76, and any number from 760 to 765 inclusive minus 6 is logically in the 750-range, we can conclude that the original two-digit number is 75. </span>

5 0

3 years ago

in 2009 a total of R36 000 was invested in two accounts. One account earned 7% annual interest and the other earned 9% .The tota

Akimi4 [234]

a = amount invested at 7%

b = amount invested at 9%

we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of a}}{\left( \cfrac{7}{100} \right)a}\implies 0.07a~\hfill \stackrel{\textit{9\% of b}}{\left( \cfrac{7}{100} \right)b}\implies 0.09b$

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.

$\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}$

5 0

3 years ago

23463.3 to 1 significant figure

mote1985 [20]

Answer:

no. Bcoz no to be drop less than 5

3 0

3 years ago

Given the following probabilities, for what P(B) are events A and B independent?

AlexFokin [52]

Answer: the value of P(B) = $\dfrac{1}{3}$ .

Step-by-step explanation:

If A and B are two independent events, then P(A and B)=P(A) x P(B) (i)

Given the following probabilities : $P(A)=\dfrac{1}{2},\ \ P(\text{A and B})=\dfrac{1}{6}$

To find : $P(B)$ such that events A and B are independent

Put all values in (i), we get

$\dfrac{1}{6}=\dfrac{1}{2}\times P(B)\\\\\Rightarrow\ P(B)=\dfrac{2}{6}\\\\\Rightarrow\ P(B)=\dfrac{1}{3}$

Hence, the value of P(B) = $\dfrac{1}{3}$ .

5 0

3 years ago

Sandy bought a 32.5 ounce package of mixed nuts for $7.I5. What was the coat per ounce.

iren [92.7K]

7.15/32.5=22
22 cents

3 0

3 years ago