Answer:
d. -24
Step-by-step explanation:
M(-1)=(-1)^2-3(-1)
M(-1)=1+3
M(-1)=4
N(-1)=-1-5
N(-1)=-6
MN(-1)=(4)(-6)
MN(-1)=-24
Answer:
- π = S/2πrh
- v = √2E/m
- b = (-7a+c)/2
- x =3y-1
Step-by-step explanation:
1.
![S = 2\pi rh\\Divide\:both\sides\:of\:the\:equation\:by\:2rh\\\\\frac{S}{2rh} = \frac{2 \pi rh}{2 rh} \\\\\pi = \frac{S}{2rh}](https://tex.z-dn.net/?f=S%20%3D%202%5Cpi%20rh%5C%5CDivide%5C%3Aboth%5Csides%5C%3Aof%5C%3Athe%5C%3Aequation%5C%3Aby%5C%3A2rh%5C%5C%5C%5C%5Cfrac%7BS%7D%7B2rh%7D%20%3D%20%5Cfrac%7B2%20%5Cpi%20rh%7D%7B2%20rh%7D%20%5C%5C%5C%5C%5Cpi%20%3D%20%5Cfrac%7BS%7D%7B2rh%7D)
2.
![E = \frac{1}{2} mv^2\\\\E = \frac{mv^2}{2} \\\\Cross\:multiply\\\\2E =mv^2\\Divide\:both\:sides\:of\:the\:equation\:by\:m\\\frac{2E}{m} = \frac{mv^2}{m} \\\\\frac{2E}{m} =v^2\\Square\:root\:both\:sides\:of\:the\:equation\\\sqrt{\frac{2E}{m} } = \sqrt{v^2} \\\\\sqrt{\frac{2E}{m} } =v\\\\v =\sqrt{\frac{2E}{m} }](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%5C%5C%5C%5CE%20%3D%20%5Cfrac%7Bmv%5E2%7D%7B2%7D%20%5C%5C%5C%5CCross%5C%3Amultiply%5C%5C%5C%5C2E%20%3Dmv%5E2%5C%5CDivide%5C%3Aboth%5C%3Asides%5C%3Aof%5C%3Athe%5C%3Aequation%5C%3Aby%5C%3Am%5C%5C%5Cfrac%7B2E%7D%7Bm%7D%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Bm%7D%20%5C%5C%5C%5C%5Cfrac%7B2E%7D%7Bm%7D%20%3Dv%5E2%5C%5CSquare%5C%3Aroot%5C%3Aboth%5C%3Asides%5C%3Aof%5C%3Athe%5C%3Aequation%5C%5C%5Csqrt%7B%5Cfrac%7B2E%7D%7Bm%7D%20%7D%20%3D%20%5Csqrt%7Bv%5E2%7D%20%5C%5C%5C%5C%5Csqrt%7B%5Cfrac%7B2E%7D%7Bm%7D%20%7D%20%3Dv%5C%5C%5C%5Cv%20%3D%5Csqrt%7B%5Cfrac%7B2E%7D%7Bm%7D%20%7D)
3.
![7a+2b =c\\Isolate\:b\\\\2b =c-7a\\2b= -7a+c\\Divide\:both\:sides\:of\:the\:equation\:by \:2\\\frac{2b}{2} =\frac{-7a+c}{2} \\\\b = \frac{-7a+c}{2}](https://tex.z-dn.net/?f=7a%2B2b%20%3Dc%5C%5CIsolate%5C%3Ab%5C%5C%5C%5C2b%20%3Dc-7a%5C%5C2b%3D%20-7a%2Bc%5C%5CDivide%5C%3Aboth%5C%3Asides%5C%3Aof%5C%3Athe%5C%3Aequation%5C%3Aby%20%5C%3A2%5C%5C%5Cfrac%7B2b%7D%7B2%7D%20%3D%5Cfrac%7B-7a%2Bc%7D%7B2%7D%20%5C%5C%5C%5Cb%20%3D%20%5Cfrac%7B-7a%2Bc%7D%7B2%7D)
4.
![y = \frac{1}{3} (x+1)\\\\y = \frac{x+1}{3}\\\\Cross\:multiply \\3y =x+1\\Isolate \:x\\3y -1=x\\\\x =3y-1](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%28x%2B1%29%5C%5C%5C%5Cy%20%3D%20%5Cfrac%7Bx%2B1%7D%7B3%7D%5C%5C%5C%5CCross%5C%3Amultiply%20%5C%5C3y%20%3Dx%2B1%5C%5CIsolate%20%5C%3Ax%5C%5C3y%20-1%3Dx%5C%5C%5C%5Cx%20%3D3y-1)
For this questions you need to memorize formula of triangle area. for question 5, just substitute the value into the formula. for Q6 you need to use pythagorean theorem formula to find the lenght of each triangle that i already seperated
Answer:
108 yards
Step-by-step explanation:
Let circle A be the circle with the 62 yard diameter
Let circle B be the circle whose diameter we are trying to solve for.
- Externally tangent circles are circles which touch each other and share a common external tangent.
- Circle A has a tangent of 62 yards and thus a radius ( half the diameter) of 31 yards.
- The distance between the centers of the 2 circles is 85 yards. If you subtract the radius ( distance from the center of the circle to its circumference) of circle A then we'll only be left with the radius of circle B.
- 85 - 31= 54 yards which is the radius of circle B
- To get the diameter: 54 x 2 = 108 yards