Answer:
All false.
Step-by-step explanation:
The ratio is 3:4.
For every 3 girls, there are 4 boys.
If a class has 7 people, 3 are girls and 4 are boys.
Hence the fraction of boys is 4/7 and girls are 3/7.
Therefore, a and b are False.
The number of boys in the class cannot be 6. If the number of boys is 6, then the number of girls must be 4.5 to maintain the 3:4 ratio, but there cannot be half a girl. So, c is also false. d) the number of pupils in the class is 12. To maintain a 3:4 ratio, the number of pupils in the class must be in multiples of 7. Hence, d is also false.
Answer:
The radius is 8 inches
Step-by-step explanation:
The volume of a cylinder is given by
V = pi r^2 h
Assuming you mean 320 pi for the volume
320 pi = pi r^2 ( 5)
320 =5pi r^2
Divide each side by 5pi
320 pi / 5pi = 5 pi r^2 / 5pi
64 = r^2
Take the square root of each side
sqrt(64) = sqrt(r^2)
8 = r
The radius is 8 inches
Answer:
1) Inequality: 110x>150+85x
2) Solution: x>6, then Nadia would need to sell more than 6 ads per week in order for Choice A to be the better choice,
<u>Solution:</u>
Be x the number of ads Nadia sells
A) Choice A: $110 for each ad she sells. She would earn:
Ea=110x
B) Choice B: A weekly salary of $150, plus $85 for each ad she sells. She would earn:
Eb=150+85x
1) Write an inequality to determine the number of ads Nadia would need to seel per week in order for Choice A to be the better choice:
Ea>Eb
110x>150+85x
2) Solution
Solving for x: Subtracting 85x both sides of the equation:
110x-85x>150+85x-85x
Subtracting:
25x>150
Dividing both sides of the equation by 25:
25x/25>150/25
Dividing:
x>6
Answer:
-1
Step-by-step explanation:
We see the given expression is equal to sin(pi/12 - 7pi/12), which is equal to sin(-pi/2), or just -1.
Angle D is 180° -75° -45° = 60°. Drawing altitude MX to segment DN divides the triangle into ΔMDX, a 30°-60°-90° triangle, and ΔMNX, a 45°-45°-90° triangle. We know the side ratios of such triangles (shortest-to-longest) are ...
... 30-60-90: 1 : √3 : 2
... 45-45-90: 1 : 1 : √2
The long side of ΔMDX is 10√3, so the other two sides are
... MX = MD(√3/2) = 15
... DX = MD(1/2) = 5√3
The short side of ΔMNX is MX = 15, so the other two sides are
... NX = MX(1) = 15
... MN = MX(√2) = 15√2
Then the perimeter of ΔDMN is ...
... P = DM + MN + NX + XD
... P = 10√3 +15√2 + 15 + 5√3
... P = 15√3 +15√2 +15 . . . . perimeter of ΔDMN
