Answer:
Pippa
Step-by-step explanation:
A square has
- parallel opposite sides
- perpendicular adjacent sides
- perpendicular diagonals
A rhombus has
- parallel opposite sides
- non-perpendicular adjacent sides
- perpendicular diagonals
Thus, we can identify the shape by comparing the slopes of the adjacent sides.
1. Draw the shape
See the graph below
2. Calculate the slope of AB
m = (y₂ - y₁)/(x₂ - x₁) = (4 - (-1))/(9 - 5) = (4 + 1)/4 = 5/4
3. Calculate the slope of BC
If BC⟂AP, its slope should be -4/5
.
m = (y₂ - y₁)/(x₂ - x₁) = (1 - 4)/(15 - 9) =-3/6 = -1/2
½ ≠ -⅘
The two lines are not perpendicular.
Pippa is right. The shape is a rhombus.
1. H+S=40
2. 19H+25S=922
From 1,
19H+19S=760
Subtract this from 2 to eliminate H,
19H+25S-19H-19S=922-760
6S=162
Solve for S, then use either equation to solve for H.
Answer How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution. ... x=15 liters 15*(.20 pure acid)=3 liters 30*(.50 pure acid)=15 liters That is 18 liters pure acid That is 45 liters solution *0.45 pure acid=18 liters.
As more pure acid is added, the concentration of acid approaches 0.
The answer is 160, multiply 12 by 4 to get 48 cats. 40 multiplied by 4 = 160.
If that makes since.
Hope this helps
<span>-Both box plots show the same interquartile range.
>Interquartile range (IQR) is computed by Q3-Q1.
For Mr. Ishimoto's class, Q3 is 35 and Q1 is 31. 35-31 = 4.
For Ms. Castillo's class, Q3 is 34 and Q1 is 30. 34-30 = 4.
</span><span>-Mr. Ishimoto had the class with the greatest number of students.
>Mr. Ishimoto had 40 students, represented by the last data point of the whiskers.
</span><span>-The smallest class size was 24 students.
>Which was Ms. Castillo's class.</span>
