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LuckyWell [14K]
3 years ago
14

Which statement correctly compares Line segment AB Which statement correctly compares Line segment AB and Line segment FD? And L

ine segment FD?
Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

AB is longer FD

Step-by-step explanation:

Given

See attachment for triangles ABC and FDE

Required

Compare line segments AB and FD

From the attachment, we have:

AC = FE --- equal line segments

The measure of angles will then be used to compare the line segments;

\angle C = 72^o

\angle F = 65^o

The longer the angle of depression, the shorter the required line segment

72 > 65 implies that AB is longer

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I need help , I don’t understand this
marta [7]
#2. First, we factor each polynomial. Then, if any terms on both the top and the bottom of the fraction match, they cancel out. So... we do just that. You end up with:

\frac{x(x-4)}{(x+9)(x-4)}

Notice there's an (x-4) on both top and bottom. So they cancel out. That leaves us with your answer of \frac{x}{(x+9)}

#3. We do the same thing as above then multiply and simplify. In the interest of space, I'll cut straight to some simplification. 

\frac{2(x+2)^{3} }{6x(x+2)} ( \frac{5}{(x-2)^{2} } )

Now we start cancelling. For the first fraction, there are 3 (x+2)'s on top and 1 on the bottom so we will cancel out the one on the bottom and leave 2 (x+2)'s on top. There are no more polynomials to cancel out so now we multiply across:

\frac{10(x+2)^{2} }{6x(x-2)^{2} }

10 and 6 share a GCF of 2 so we divide both of those by 2. This leaves us with the final answer of:

\frac{5(x+2)^{2} }{3x(x-2)^{2} }

#4. This equation introduces division and because of it, we must flip the second fraction to make the division sign into a multiplication symbol. Again for space, I'll flip the fraction and simplify in one step. 

\frac{3(x+2)(x-2)}{(x+4)(x-2)} ( \frac{x+4}{6(x+3)})

Now we do our cancelling. First fraction has (x - 2) in the top and bottom. They're gone. The first fraction has a (x + 4) on the bottom and the second fraction has one on the top. Those will also cancel. This leaves you with:

\frac{3(x+2)}{6(x+3)}

3 and 6 share a GCF of 3 so we divide both numbers by this. This leaves you with your final answer:

\frac{x+2}{2(x+3)}

#5. We are adding so we first factor both fractions and see what we need to multiply by to make the denominators the same. I'll do the former first. (10 - x) and (x - 10) are not the same so we multiply the first equation (top and bottom) by (x - 10) and the second equation by (10 - x). Because they will now have the same denominator we can combine them already. This gives us:

\frac{(3+2x)(x-10)+(13+x)(10-x)}{(10-x)(x-10)}

Now we FOIL each to expand and then simplify by combining like terms. Again for space, I'm just showing the result of this; you end up with:

\frac{x^{2}-20x+100}{(10-x)(x-10)}

Now we factor the top. This gives you 2 (x - 10)'s on top and one on bottom. So we just leave one on the top and cancel the bottom one out. This leaves you with your answer:

\frac{x+10}{10-x}

#6. Same process for this one so I won't repeat. I'll just show the work.

\frac{3}{(x-3)(x+2)} +  \frac{2}{(x-3)(x-2)} becomes

\frac{3(x-2) + 2(x+2)}{(x-3)(x+2)(x-2)} which equals

\frac{3x - 6 + 2x + 4}{(x-3)(x+2)(x-2)} giving you the final answer

\frac{5x - 2}{(x-3)(x+2)(x-2)}

#7. For this question we find the least common denominator to make the denominators match. For 5, x, and 2x, the LCD is 10x. So we multiply top and bottom of each fraction by what would make the bottom equal 10x. This rewrites the fraction as:

\frac{3x}{5} ( \frac{2x}{2x}) * ( \frac{5}{x}( \frac{10}{10}) -  \frac{5}{2x} ( \frac{5}{5}))

Simplify to get:

\frac{3x}{5}  * ( \frac{25}{10x})

After simplifying again, you end up with your final answer: 

\frac{3}{2}




8 0
3 years ago
Of the Students that attended Roosevelt Elementary 6/8 of the school play a sport. Of the students are playing a sport 3/5 of th
soldier1979 [14.2K]

Answer:

\frac{9}{20} students.

Step-by-step explanation:

We have been given that of the students that attended Roosevelt Elementary 6/8 of the school play a sport. Of the students are playing a sport 3/5 of the students are also involved in the drama club.

To find the students are both play a sport and a part of the drama club, we need to find 3/5 part of 6/8 as:

\text{Students who play a sport and a part of the drama club}=\frac{6}{8}\times \frac{3}{5}

\text{Students who play a sport and a part of the drama club}=\frac{3}{4}\times \frac{3}{5}

\text{Students who play a sport and a part of the drama club}=\frac{3\times 3}{4\times 5}

\text{Students who play a sport and a part of the drama club}=\frac{9}{20}

Therefore, \frac{9}{20} students play sport and part of the drama club.

5 0
3 years ago
Two consecutive odd integers have a sum of 32. Find the integers.
bezimeni [28]
Let
x = first consecutive odd
x + 2 = second consecutive odd

Based on the problem, we equate
x + (x + 2) = 32
Solving for x,
2x + 2 = 32
2x = 32 - 2
2x = 30
x = 30/2
x = 15
and x + 2 = 15 + 2 = 17

Therefore, the integers are 15 and 17.
6 0
3 years ago
|2x-3|=2 please answer this by Wednesday. I have no idea how to solve this. I wasn't paying attention
Sholpan [36]

Answer:

x=2.5

Step-by-step explanation:

3 0
3 years ago
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