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Kruka [31]
3 years ago
13

Find the distance of each segment AB and CD. You may leave in radical form. A (3, 4) B(1, 0) and C (-1, 2) D (-2, 5)

Mathematics
2 answers:
Whitepunk [10]3 years ago
5 0

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AB

\sqrt{(1-3)^2+(0-4)^2}

\sqrt{4+16}

\sqrt{20}

2\sqrt5

CD

\sqrt{(-2-(-1))^2+(5-2)^2}

\sqrt{1+9

\sqrt{10}

Hope this helps.

頑張って!

nevsk [136]3 years ago
4 0

Answer:

Step-by-step explanation:

se the graph to determine the input values that

correspond with f(x) = 1.

O x=4

O x= 1 and x = 4

O x= -7 and x = 4

O x= -7 and x = 2

6.

(-6, 4)

4

(1,4) w

2

(-7, 1)

(2, 1) x w

2

4

-8/ -6 -4 -2 e

-2

-4thrrt

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scoundrel [369]
Ten is to the 6th power so there are 6 zeroes. 
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3 years ago
Simplify 7x²+3x-5x²-10x
solmaris [256]

Answer:

2x²-7x

Step-by-step explanation:

7x²-5x²=2x²

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3 0
3 years ago
Can someone help me with this??
dusya [7]

Answer:

Carlos and Pamela drove 120 miles on the first day, 240 miles on the second day, and 290 miles on the third day.

Step-by-step explanation:

Let x be the number of miles driven on the first day.

Then they drove twice as many miles, or 2x on the second day

and 50 miles more than the second day's, so 2x + 50

The total is 650 across all three days, so we'll take the sum.

x + 2x + (2x + 50) = 650

Combine like terms on the left

5x + 50 = 650

Subtract 50 on both sides

5x = 600

Divide by 5 on both sides

x = 120

Check work:

120 + 2(120) + 2(120) + 50 = 650

120 + 240 + 240 + 50 = 650

600 + 50 = 650

650 = 650

So they drove 120 miles on the first day

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4 0
3 years ago
A population is estimated to have a standard deviation of 9. We want to estimate the population mean within 2, with a 99% level
cupoosta [38]

Answer:

The sample required is  n = 135

Step-by-step explanation:

From the question we are told that

     The  standard deviation is  \sigma = 9

      The margin of error is E =  2

     

Given that the confidence level is  99%  then the level of  significance is mathematically evaluated as

         \alpha =  100-99

        \alpha =  1 \%

        \alpha =  0.01

Next we will obtain the critical value  \frac{\alpha }{2} from the normal distribution table(reference  math dot armstrong dot edu) , the value is  

             Z_{\frac{\alpha }{2} } =  Z_{\frac{0.05 }{2} } =  2.58

The  sample size is mathematically represented as

          n = [ \frac{Z_{\frac{\alpha }{2} } *  \sigma }{E} ]^2

substituting values

           n = [ \frac{ 2.58 *  9 }{2} ]^2

            n = 135

3 0
3 years ago
Jessica scores 1825 on the sat. ashley scores 28 on the act assuming that both tests measure the same thing who has the higher s
SOVA2 [1]
<span>With 1026 being the mean score on the SAT and StDev of 209, Jessica has a score of 799/209 z-scores above the mean. Her z-score would be 3.823. The mean of the ACT and StDev are 20.8 and 4.8, respectively. A 28 ACT score would be 7.2/4.8 z-scores above the mean, for a z = 1.500. This means that Jessica has the higher z-score.</span>
5 0
3 years ago
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