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Reptile [31]
3 years ago
11

What is the answer with this math

Mathematics
2 answers:
xenn [34]3 years ago
3 0
Find the least common denominator of 2 and 3, which is 6. Because you had to multiply 2 by 3 to get 6, you have to multiply 1 by 3, you then get 3/6. Because you had to multiply 3 by 2 to get 6, you have to multiply 1 by 2, you then get 2/6. Now, you can subtract 3/6 - 2/6, which equals 1/6. Hope this helps!
sergiy2304 [10]3 years ago
3 0
The Answer Should Be 1/6
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Molly wrote messages to her friends on the sidewalk. She used 1/4 of a piece of chalk that was 2 3/5 inches long. How many inche
IRINA_888 [86]

Answer:3/5 inches!!!!!!!!

5 0
2 years ago
Read 2 more answers
The base of a 15 foot long guy wire is located 5 ft from the telephone pole it is anchoring. how high up the pole does the guy w
Andrei [34K]

{5}^{2}  +  {h}^{2}  =  {15}^{2}
h = sqrt
sqrt{200}

6 0
3 years ago
Use the Law of Cosines to find angle round to the nearest tenth if necessary).
ad-work [718]

Answer:

Step-by-step explanation:

3a)

find side HP  , let's set side HP = c in the law of cosines formula

then

c = sq rt [ 23^2 + 12^2 - 2*12*23*cos(117) ]

c = sq rt [ 529 + 144 - (-250.6027559) ]

c = sq rt [ 673 + 250.6027559 ]

c = sq rt [ 923.602755 ]

c = 30.39

then

HP = 30.4  ( rounded to nearest 10th )

3b)

let PR = c, then law of cosines is (we need angle Q also 180=24+97+Q is 59°

c = sq rt [ 9^2 + 22^2 - 2*9*22*cos(59) ]

c = sq rt [ 81 + 484 -396*0.5150380 ]

c = sq rt [ 565 - 203.955 ]

c = sq rt [ 391.04495]

c = 19.7748

PR = 19.8  ( rounded to nearest 10th )

3c)

they want us to find ∠B,   in the formula for cosines ∠B is really ∠C, it's a bit confusing.. I'm sure they did that on purpose to mess people up.  Kind of underhanded  :/    Tell the teacher that you're "appalled" at the deviousness of the test  :/   stage a protest against math that teaches treachery  :P  storm the principals/ dean's  office with paper airplanes and pictures of anime   :DDD  okay back to math.

a=28

b=17

c=15

since we want angle B then the formula looks like

B= arcCos ( a^2 + c^2 - b^2 / 2*a*c)  so we'll avoid messing up the which side is which ... by setting up the formula this way   .. got it?

B = arcCos [28^2 + 15^2 - 17^2 / 2*28*15]

B = arcCos[784 + 225 - 289 /840 ]

B = arcCos[720/ 840]

B = arcCos[0.85714285]

B = 31.0027 °

that it came out so close to an integer I'm pretty sure it's right

∠B = 31 .0 °  ( rounded to the nearest 10th )

3d)

find ∠A

set it up carefully again, they are asking us to move the variable around again

a = 18

b= 28

c= 12

then the formula looks like

∠A = arcCos[ b^2 + c^23 - a^2 / 2*b*c ]

∠A = arcCos[28^2 + 12^2 - 18^2 /2*28*12 ]

∠A = arcCos[784 + 144 -324 / 672 ]

∠A = arcCos[604/672]

∠A = arcCos[0.8988095]

∠A =25.99797 °

∠A =26.0 °  ( rounded to nearest 10th)

7 0
3 years ago
The number n when increased
Olin [163]

Answer:

Number n increased by 18 = n +18

Number is 2 less than 5 times n= 5n -2

They are same =>  n +18 = 5n -2

Option B

[ n + 18 = 5n -2

18 +2 =5n - n

20 = 4n

n = 5

Lets check 5+18 = 5*5 - 2

                 23 = 23]

6 0
3 years ago
A principal gathered data about the distance, in miles, that his teachers and bus drivers live from school. The box plots below
Gekata [30.6K]

Answer:

We choose C

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.  Please have a look at the attached photo

Basically, interquartile range represents the width or "dispersion" of the set. [1] The interquartile range is determined by the difference between the top quartile (25% highest) and lower quartile (25% lowest) point of the data set.

From the picture, we can find that:

  • The interquartile range of the bus drivers is: 20 -10 = 10
  • The interquartile range of the teachers is: 30 -15 = 15

So the interquartile range of the distances for the bus drivers is 5 miles less than the interquartile range of the distances for the teachers.

We choose C

4 0
3 years ago
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