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atroni [7]
3 years ago
12

What is the slope of the line containing the point (34,-27) and (14, 37)

Mathematics
2 answers:
Vadim26 [7]3 years ago
8 0
The slope of the line is 64/-24. That reduces to 8/-3.

Use the formula, Y2-Y1/X2-X1.
Korolek [52]2 years ago
3 0

Step-by-step explanation:

where slope is equal to y2-y1/x2-x1

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4(x+8)-2(x-7)=46 x=?
Scorpion4ik [409]

Answer:

x = 0/2

x = 0

Step-by-step explanation:

4(x+8)-2(x-7) = 46

Expanding the brackets

4x +32 - 2x +14 = 46

2x + 46 = 46

2x = 46-46

2x = 0

x = 0/2

x = 0

5 0
3 years ago
What number completes the sequence below? enter your answer below
Readme [11.4K]

Answer:

18 + 5 = 23

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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
Assume that the mean length of a human pregnancy is 269 days. If​ 95% of all human pregnancies last between 249 and 289 ​days, w
o-na [289]

Answer:93% not sure

Step-by-step explanation:

7 0
2 years ago
A baker made two cakes of the same size. • At the end of the day, there was 23 of a chocolate cake left. • There was 56 of a str
Murrr4er [49]
This is the concept of algebra, To solve the question  we proceed as follows;
Number of chocolate cake left=23 
The number of pieces that was left after the cake was divided into 2 equal pieces will be:
23*2
=46

Number of strawberry cake left = 56
The number of pieces that was left after the cake was divided into 3 equal pieces will be:
56*3
=168

Comparing the two fraction, the flavor that had larger pieces was strawberry cake;
This cake was larger compared to chocolate by:
168-46
=122
The answer is Strawberry by 122 of a cake
5 0
3 years ago
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