136 - (-50)= 186
186 is the difference between the too temperatures.
Answer:
x
≤
13
Step-by-step explanation:
hope this helps
To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:
![\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20ax%5E2%2Bbx%2Bc%3D0%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5Cend%7Bgathered%7D)
For the given function:
![x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-10%29%5Cpm%5Csqrt%5B%5D%7B%28-10%29%5E2-4%282%29%28-3%29%7D%7D%7B2%282%29%7D)
![x=\frac{10\pm\sqrt[]{100+24}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B100%2B24%7D%7D%7B4%7D)
![\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B124%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5Ccdot2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5E2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%3D%5Cfrac%7B10%7D%7B4%7D%2B%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_1%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_2%3D%5Cfrac%7B10%7D%7B4%7D-%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_2%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Then, the zeros of the given quadratic function are:
![\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5C%5C%20%20%5C%5C%20x_%7B%7D%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Answer: Third option
Answer:
126 degrees
Step-by-step explanation:
Triangle: y = 180 - (5x + 4x)
Straight angle: y = 180 - (4x + 30)
180 - (5x + 4x) = 180 - (4x + 30)
5x + 4x = 4x + 30
5x = 30
x = 6
Angle:
y = 180 - (5x + 4x)
y = 180 - 9x
y = 180 - 9(6)
y = 126
Answer:
5.83 inches
Step-by-step explanation:
The card is a rectangular shaped object, drawing a diagonal from one corner to the other, divides the rectangle into two right angled triangles with the diagonals as the hypothenuse.
Using Pythagoras theorem.
c^2 = a^2 + b^2
Given;
Side a = 3
Side b = 5
Substituting the given values;
c^2 = 3^2 + 5^2
c = √(9+25)
c = √34
c = 5.83 inches
