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3 years ago

16.Can you use the SAS Postulate, the AAS Theorem, or both to prove the triangles congruent?

Mathematics

2 answers:

Elza [17]3 years ago

8 0

C I just had this question on a post test..

professor190 [17]3 years ago

6 0

The answer is C

You are already given one angle and one side. Using the vertical angles postulate we know that the two verticle angles are congruent, satisfying the AAS criteria

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1/7x1/7x1/7= 3/7 true or false

Lostsunrise [7]

Answer:

The answer is false

3 0

3 years ago

Read 2 more answers

Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?

garik1379 [7]

Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is $z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$ .

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be $z (x) = \cos x$ the base formula, where $x$ is measured in sexagesimal degrees. This expression must be transformed by using the following data:

$T = 180^{\circ}$ (Period)

$z_{min} = -4$ (Minimum)

$z_{max} = 5$ (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of $2\pi$ radians. In addition, the following considerations must be taken into account for transformations:

1) $x$ must be replaced by $\frac{2\pi\cdot x}{180^{\circ}}$ . (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

$\Delta z = \frac{z_{max}-z_{min}}{2}$

$\Delta z = \frac{5+4}{2}$

$\Delta z = \frac{9}{2}$

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

$z_{m} = \frac{z_{min}+z_{max}}{2}$

$z_{m} = \frac{1}{2}$

The new function is:

$z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)$

Given that $z_{m} = \frac{1}{2}$ , $\Delta z = \frac{9}{2}$ and $T = 180^{\circ}$ , the outcome is:

$z'(x) = \frac{1}{2} + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)$

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

8 0

3 years ago

In a store, a sample of 60 pens was selected from a box of 1,200 pens. Of these 60, 4 pens did not work. About how many pens cou

Lapatulllka [165]

Ok so if 1200 total pens and 60 were taken that's a ratio of 1200/60=20

out of 60pens 4 didn't work, again ratio is 4/60=1/15 so about 1/15 of the pens didn't work.

so we have 1200pens * 1/15broken ratio = 80 pens or we can say

20*4=80pens won't work from 20 groups of 60 and 4 out of 60 not working is 20*4=80 not working from total batch of 1200.

hope this helps you some! thank you!!!!!!!!!

3 0

3 years ago

How do I solve this equation?

frosja888 [35]

Answer:

x=-3/11

Step-by-step explanation:

(x+3)/5 +2x=0

Multiply both sides by 5 to get rid of the fraction

5*( x+3)/5 +5*2x=0*5

x+3 + 10x =0

Combine like terms

11x +3 = 0

Subtract 3 from both sides

11x+3-3 =0-3

11x = -3

Divide by 11

11x/11 = -3/11

x = -3/11

4 0

3 years ago

WILL MARK BRANLIEST! find of the volume of the prism round to the nearest tenth if necessary

vodomira [7]

Answer:

The answer is A. 204.7

Step-by-step explanation:

5 0

3 years ago