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Advocard [28]
3 years ago
9

What theorem can be used to prove the triangles are congruent

Mathematics
1 answer:
natta225 [31]3 years ago
7 0

Answer:

Side-Angle-Side is a rule used to prove whether a given set of triangles are congruent. The SAS rule states that: If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.

Step-by-step explanation:

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What is the point of intersection of these two lines? -4x + y = 8 2x - y = 4 Question 5 options: (-6, -16) (-2, -8) (2, 0) (−23,
Snezhnost [94]

Answer:

(-6,-16)

Step-by-step explanation:

4 0
3 years ago
Graphing a line given its slope and y-intercept
kifflom [539]

Answer:

  see below for a graph

Step-by-step explanation:

One point can be plotted at the y-intercept: (0, -2). Since the slope tells you the line drops 3 units for each 2 units to the right, the point (2, -5) will be another point on the line. The graph will go through those two points.

5 0
4 years ago
Eight more than three times a number is equal to the sum of the number and 10. find the number
Afina-wow [57]

Answer:

8+3x=x+10

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x=1

The number is 1

6 0
3 years ago
The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p
VLD [36.1K]

Answer:

3.13

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

\chi^2_{1-\frac{\alpha}{2}}= 51.770

Chi square critical value for upper tail =

\chi^2_{\frac{\alpha}{2}}= 81.085

80% confidence interval:

\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}

Putting values, we get,

=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

8 0
3 years ago
System practice problems
nydimaria [60]

Looking at the first system of equations,

16x - 10y = 10

-8x - 6y = 6

If we multiply both sides of the second equation by 2, the coefficient of x is exactly the negative of the coefficient of x in the first equation.

-8x - 6y = 6

⇒   2 (-8x - 6y) = 2 (6)

⇒   -16x - 12y = 12

By combining this new equation with the first one, we can eliminate x and solve for y :

(16x - 10y) + (-16x - 12y) = 10 + 12

⇒   -22y = 22

⇒   y = -1

Then we just solve for x by replacing y in either equation.

16x - 10y = 10

⇒   16x - 10 (-1) = 10

⇒   16x + 10 = 10

⇒   16x = 0

⇒   x = 0

The main idea behind elimination is combining the given equations in just the right amount so that one of the variables disappears. The "right amount" involves using the LCM of the coefficients of a given variable. In this example, the x-coefficients had LCM(8, 16) = 16, so we only had to scale one of the equations (the one with -8x) to cancel all the x terms.

If we wanted to eliminate y first instead, we first note that LCM(6, 10) = 30. To get 30 as a coefficient on y, in the first equation we would have multiplied by 3:

16x - 10y = 10

⇒   3 (16x - 10y) = 3 (10)

⇒   48x - 30y = 30

And in the second equation, we would have multiplied by -5 (negative so that upon combining the equations, we end up with -30y + 30y = 0):

-8x - 6y = 6

⇒   -5 (-8x - 6y) = -5 (6)

⇒   40x + 30y = -30

Now combining the two scaled equations gives

(48x - 30y) + (40x + 30y) = 30 + (-30)

⇒   88x = 0

⇒   x = 0

We then solve for y :

16x - 10y = 10

⇒   -10y = 10

⇒   y = -1

so we end up with the same solution as before.

8 0
3 years ago
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