Answer:
I do it's option B your welcome!
-x+2 > 1
-x+2+x > 1+x .... add x to both sides
2 > 1+x
x+1 < 2
x+1-1 < 2-1 ... subtract 1 from both sides
x < 1
After solving for x, we get x < 1
To graph this, plot an open circle at 1 on the number line and shade to the left of this value. The open circle indicates that 1 is not part of the solution set.
If your teacher requires you to graph this on an xy grid, then draw a vertical line through 1 on the x axis. Make this vertical line a dashed line. Then shade the entire region to the left of this dashed line. Any point in this shaded region will have an x coordinate that is less than 1. The dashed line acts like the open circle. The dashed line tells the reader "any point on this dashed line is not part of the solution set"
<span> Answer: 1 One way to determine the greatest common factor is to find all the factors of the numbers and compare them.
The factors of 26 are 1, 2, 13, and 26.
The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
The factors of 75 are 1, 3, 5, 15, 25, and 75.
The only common factor is 1. Therefore, the greatest common factor is 1</span>
<span>
The greatest common factor can also be calculated by identifying the common prime factors and multiplying them together.
The prime factors of 26 are 2 and 13.
The prime factors of 48 are 2, 2, 2, 2, and 3.
The prime factors of 75 are 3, 5, and 5.
The are no prime factors in common, so the greatest common factor is 1.
Another way to approach this is to look at the differences between the numbers. The difference between 26 and 48 is 22. The difference between 48 and 75 is 27. The greatest common factor of two or more numbers cannot be larger than the smallest difference between the numbers. The greatest common factor of 26, 48, and 75 must also be a factor of the differences between the numbers. So, the greatest common factor of 22 and 27 is also the greatest common factor of 26, 48, and 75. The greatest common factor of 22 and 27 is 1, so the greatest common factor of 26, 48, and 75 is also 1.
</span>
First we dra a triangle:
To prove that the triangles are similar we have to do the following:
Considet triangles ABC and ACD, in this case we notice that angles ACB and ADC are equal to 90°, hence they are congruent. Furthermore angles CAD and CAB are also congruent, this means that the remaining angle in both triangles will also be congruent, therefore by the AA postulate for similarity we conclude that:
Now consider triangles ABC and BCD, in this case we notice that angles ACB and BDC are congruent since they are both equal to 90°. Furthermore angles ABC and DBC are also congruent, this means that the remaining angle in both triangles will, once again, be congruent. Hence by the AA postulate we conclude that:
With this we conclude that traingles BCD and ACD are both similar to triangle ABC, and by the transitivity property of similarity we conclude that:
Now that we know that both triangles are similar we can use the following proportion:
this comes from the fact that the ratios should be the same in similar triangles.
From this equation we can find h:
![\begin{gathered} \frac{h}{x}=\frac{y}{h} \\ h^2=xy \\ h=\sqrt[]{xy} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bh%7D%7Bx%7D%3D%5Cfrac%7By%7D%7Bh%7D%20%5C%5C%20h%5E2%3Dxy%20%5C%5C%20h%3D%5Csqrt%5B%5D%7Bxy%7D%20%5Cend%7Bgathered%7D)
Plugging the values we have for x and y we have that h (that is the segment CD) has length:
![\begin{gathered} h=\sqrt[]{8\cdot5} \\ =\sqrt[]{40} \\ =\sqrt[]{4\cdot10} \\ =2\sqrt[]{10} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%3D%5Csqrt%5B%5D%7B8%5Ccdot5%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B40%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B4%5Ccdot10%7D%20%5C%5C%20%3D2%5Csqrt%5B%5D%7B10%7D%20%5Cend%7Bgathered%7D)
Therefore, the length of segment CD is:
![CD=2\sqrt[]{10}](https://tex.z-dn.net/?f=CD%3D2%5Csqrt%5B%5D%7B10%7D)
Answer:
E
Step-by-step explanation:
I don't understand in this specific context what is evaluated means but looking over the options it seems that 0 to the negative 1/2 power seems pretty unlikely . the only reason I don't say 0 to the negative 2nd power is because 0 to the negative 2nd power seems plausible but it doesn't seem like 0 to the negative 1/2 power would ever become a thing
