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DIA [1.3K]
2 years ago
14

What additional information is needed to prove the triangles are congruent by SAS ?

Mathematics
1 answer:
Wittaler [7]2 years ago
5 0

Answer: Its EB≅CB

Step-by-step explanation: I took the test

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3 times 80,000,00 equal what
RoseWind [281]

Answer:

240000000

Step-by-step explanation:

3*80000000

=240000000

7 0
3 years ago
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Please answer! I crossed out the ones you don’t have to complete.
Nina [5.8K]

Answer:

1. Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: \mathbf{5^3a^2b^4c}

5.  x^-6 = \frac{1}{x^6}

6. 5^{-3}.3^{-1}=\frac{1}{5^3.3^1}

7. a^{-3}b^0c^4=\frac{c^4}{a^3}

Step-by-step explanation:

Question 1:

We need to rewrite the expression using exponents

5.a.b.b.5.c.a.b.5.b

We will first combine the like terms

5.5.5.a.a.b.b.b.b.c

Now, if we have 5.5.5 we can write it in exponent as: =5^{1+1+1}=5^3

a.a as a^{1+1}=a^2

b.b.b.b as: b^{1+1+1+1}=b^4

So, our result will be:

5^3a^2b^4c

Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: \mathbf{5^3a^2b^4c}

Question:

Rewrite using positive exponent:

The rule used here will be: a^{-1}=\frac{1}{a^1} which states that if we need to make exponent positive, we will take it to the denominator.

Applying thee above rule for getting the answers:

5) x^{-6} = \frac{1}{x^6}

6) 5^{-3}.3^{-1}=\frac{1}{5^3.3^1}

7) a^{-3}b^0c^4=\frac{b^0c^4}{a^3}

We know that b^0=1 so, we get

a^{-3}b^0c^4=\frac{b^0c^4}{a^3}=\frac{c^4}{a^3}

4 0
3 years ago
Please answer :) Easy question
lutik1710 [3]
4. ^

5. A slope is a line nevertheless and it keeps changing, while the rise over run is also a slope. This means that the only correct answer can be a 2 dimensional figure.

7 0
3 years ago
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Write an equation of a line passing
Makovka662 [10]

Answer:

slope of line pass throught jk is 0.5 slope

since line perpendicular to jk then x × 0.5= -1 so slope of line perpendicular is -2

then eq of line is y= -2x -15

6 0
3 years ago
Determine which equation is parallel to the line JK and which is perpendicular to like JK
sattari [20]

Answer:

If the equation of JK is y = mx + c, then y = mx + c' is the equation of the straight line parallel to above and y = - \frac{1}{m}x + b will be perpendicular to that.

Step-by-step explanation:

Let us assume that the line JK has the equation in slope-intercept form as  

y = mx + c ............. (1)

Therefore, the equation has slope = m

Now, any straight line having an equation with slope m will be parallel to line JK.

So, the equation of the straight line which is parallel to equation (1) will be  

y = mx + c', where, c' is any real constant.

Now, let us assume that another straight line having equation  

y = nx + a is perpendicular to the line JK i.e. equation (1).

Now, we know if two lines are perpendicular to each other then the product of their slopes will be - 1.

So, mn = - 1

⇒ n = - \frac{1}{m}

Therefore, the equation of a straight line which is perpendicular to equation (1) will be y = - \frac{1}{m}x + b where b is any real constant. (Answer)

3 0
3 years ago
Read 2 more answers
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