Let's solve your equation step-by-step.
5=(−4)(8)+b
Step 1: Simplify both sides of the equation.
5=(−4)(8)+b
5=−32+b
5=b−32
Step 2: Flip the equation.
b−32=5
Step 3: Add 32 to both sides.
b−32+32=5+32
b=37
Any line can be expressed in the form y=mx+b where m is the slope and b is y intercept.
Two lines can either be parallel ,overlap or meet at one point .Let us look at different cases :
1)When two lines are parallel they do not intersect at any point and hence the system of equations have no solution.
2) When two lines overlap each other then the two lines touch each other at infinite number of points and we say the system of equations have infinite solutions.
3) When two lines intersect each other at one point we say the system of equation has one solution.
Part A:
The given lines are intersecting at one point so we have one solution.
Part B:
The point of intersection is the solution to the system of equations .In the graph the point of intersection of the lines is (4,4)
Solution is (4,4)
I would assume it is 0% for the minimum as there is a chance that rangers didn’t see anything and 65% for the highest as that is all the people that saw giraffe. I am really unsure though.
My second answer is minimum is 55.25% and the maximum is 65%. But once again I ain’t sure and logically I would go for my first answer
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
