Answer:
xº=71º
Step-by-step explanation:
Take the triangle BDH
Once the sum of interior angles of a triangle is 180º:
40º+31º+yº=180º
71º+yº=180º
yº=109º
Once yº and xº are supplementary angles (add up to 180º):
yº+xº=180º
109º+xº=180º
xº=71º
Answer:
1 hour
Step-by-step explanation:
Hello, let's say that her departure trip takes t in minutes, as her return speed is 3 times her departure speed, she took t/3 for the return and we know that this 40 minutes less, so we can write.
t/3=t-40
We can multiply by 3
t = 3t -40*3 = 3t - 120
This is equivalent to
3t -120 = t
We subtract t
2t-120 = 0
2t = 120
We divide by 2
t = 120/2 = 60
So this is 60 minutes = 1 hour.
Thank you.
There an old algorythmic method similar to long division:-
7 . 4 8 3
-----------------
) 56. 00 00 00
7 ) 49
) ---
144 ) 7 00
) 576
----
148 8 ) 12400
) 11904
-------
) 496 00
1496 3 ) 44889
THis gives the square root as 7.48 to the nearest hundredth
Dimensions are length 20 meter and width 14 meter
<em><u>Solution:</u></em>
Let "a" be the length of rectangle
Let "b" be the width of rectangle
Given that,
<em><u>A rectangle has width that is 6 meters less than the length</u></em>
Width = length - 6
b = a - 6
The area of the rectangle is 280 square meters
<em><u>The area of the rectangle is given by formula:</u></em>
![Area = length \times width](https://tex.z-dn.net/?f=Area%20%3D%20length%20%5Ctimes%20width)
<em><u>Substituting the values we get,</u></em>
![Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0](https://tex.z-dn.net/?f=Area%20%3D%20a%20%5Ctimes%20%28a-6%29%5C%5C%5C%5C280%20%3D%20a%5E2-6a%5C%5C%5C%5Ca%5E2-6a%20-280%3D0)
<em><u>Solve the above equation by quadratic formula</u></em>
![\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3Aa%5C%3Aquadratic%5C%3Aequation%5C%3Aof%5C%3Athe%5C%3Aform%5C%3A%7Dax%5E2%2Bbx%2Bc%3D0%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7D%5C%5C%5C%5C%5Cquad%20x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7D%5Cquad%20a%3D1%2C%5C%3Ab%3D-6%2C%5C%3Ac%3D-280%3A%5Cquad%20a_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-%5Cleft%28-6%5Cright%29%5Cpm%20%5Csqrt%7B%5Cleft%28-6%5Cright%29%5E2-4%5Ccdot%20%5C%3A1%5Cleft%28-280%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A1%7D)
![a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7B6%20%5Cpm%20%5Csqrt%7B36%2B1120%7D%7D%7B2%7D%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B6%20%5Cpm%20%5Csqrt%7B1156%7D%7D%7B2%7D%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B6%20%5Cpm%2034%7D%7B2%7D%5C%5C%5C%5CThus%5C%20we%5C%20have%5C%20two%5C%20solutions%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B6%2B34%7D%7B2%7D%20%5Ctext%7B%20or%20%7D%20a%20%3D%20%5Cfrac%7B6-34%7D%7B2%7D%5C%5C%5C%5Ca%20%3D%2020%20%5Ctext%7B%20or%20%7D%20a%20%3D%20-14)
Since, length cannot be negative, ignore a = -14
<em><u>Thus solution of length is a = 20</u></em>
Therefore,
width = length - 6
width = 20 - 6 = 14
Thus dimensions are length 20 meter and width 14 meter
Answer:
Step-by-step explanation:
Discussion
The first thing you have to do is convert km to miles
Formula
1 mile = 1.6 km
x mile = 40000 km Cross Multiply
Solution
1.6*x = 1 * 40000 Divide by 1.6
x = 40000/1.6
x = 25000 miles
Answer Part 1
The distance covered is 25000 miles
Discussion Part 2
That's the distance covered in a day. Now we need to convert that to speed. The time (in hours) for 1 day is 24 hours.
Givens
d = 25000 miles
t = 24 hours
s = ?
Solution
s = d/t
s = 25000 / 24
s = <u>1041.66 </u> miles per hour
Answer: 1041.66 is what is to be entered as your answer