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meriva
2 years ago
8

Solve for X. Will give brainliest!

Mathematics
2 answers:
shtirl [24]2 years ago
6 0

Answer:

x=2

Step-by-step explanation:

The length from B to D appears to be the same length as from D to E and A to B and twice the length of B to C.

irga5000 [103]2 years ago
3 0

Answer:

2

Step-by-step explanation:

Since triangles ABC and ADE are similar by AA (they share angle A and a right angle), their ratio of sides must be equal. Therefore:

\dfrac{1}{2}=\dfrac{2}{x+2}

Multiply both sides by 2:

1=\dfrac{4}{x+2}

Multiply both sides by x+2:

x+2=4

Subtract 2 from both sides:

x=2

Hope this helps!

You might be interested in
The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

Now Dbar = Bbar - Abar = 489 - 514 = -25

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

                            = \frac{-25 - 0}{\frac{37.05}{\sqrt{12} } } follows t_1_1   = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0
3 years ago
What is the rate of heat transfer required to melt 1-ton of ice at 32 F in 24 hours?
LiRa [457]

Answer:

3865.74 J/s

Step-by-step explanation:

mass of ice, m = 1 ton = 1000 kg

time , t = 24 hours

latent heat of fusion of ice, L  = 334000 J/kg

Heat required to melt, H = m x L

where, m is the mass of ice and L be the latent heat of fusion

So, H = 1000 x 334000 = 334 xx 10^6 J

Rate of heat transfer = heat / time = \frac{334\times 10^{6}}{86400}

Rate of heat transfer = 3865.74 J/s

thus, the rate of heat transfer is 3865.74 J/s.

8 0
3 years ago
The following graph represents the distance a commercial airplane travels over time, at cruising speed and an altitude of 35,000
ANEK [815]

Answer:

see the explanation

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form k=\frac{y}{x} or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Let

x ----> the time in hours

y ----> the distance in miles

<em>Find the value of k</em>

For the point (4,2268)

k=\frac{2,268}{4}=567\ mph

The slope represent the speed of the airplane

so

The linear equation is

y=567x

Part 1 :

The point (0,0) represents the starting point of the aircraft, when the time and distance are equal to zero. The cruising starts when time t = 0.

Part 2 :

The  point  (4, 2268) represents the plane after 4 hours of cruise , and shows it has traveled a distance of 2268 miles after 4 hours

3 0
3 years ago
OeowowowkwiHELP ASaPieisowowiiwiwiwiiwiwisnsnsnxxj
puteri [66]

Answer:

5

Step-by-step explanation:

40÷[20-4*(7-4)]

Start with the inner most parentheses

40÷[20-4*(3)]

Then the brackets, multiply first

40÷[20-12]

Then subtract

40÷[8]

We are now left with the division

5

6 0
3 years ago
PLS HALP ASAP =w=!!!! (will mark u teh brainliest if you explain the problem properly and clearly, 20 pts uwu!)
Oksanka [162]
N = 3 i = 2
2 x 2 = 4
3 x 3 = 9
4 + 9 = 13
13 = area
7 0
2 years ago
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