An equation whose variables are polar coordinates is called a polar equation. These equation are characterized by an r as a function an angle. Polar equations can be written in rectangular coordinates by certain relationships. An example of a polar equation would be r = 2sin∅.
Answer:
it is 234
Step-by-step explanation:
Answer:
50 gluten-free cupcakes and 100 regular cupcakes.
Step-by-step explanation:
Let's define the variables:
R = number of regular cupcakes sold
G = number of gluten-free cupcakes sold
The total amount of money raised then is:
M = R*$2.00 + G*$3.00
We also know that:
The number of regular cupcakes sold was 2 times the number of gluten-free cupcakes sold.
then:
R = 2*G
And we also know that the amount of money raised is $350
Then we have the equations:
R = 2*G
R*$2.00 + G*$3.00 = $350
We can replace the first equation into the second one, so we have only one variable:
(2*G)*$2.00 + G*$3.00 = $350
Now we can solve this for G.
G*$4.00 + G*$3.00 = $350
G*$7.00 = $350
G = $350/$7.00 = 50
G = 50
50 gluten-free cupcakes where sold.
And using the equation:
R = 2*G = 2*50 = 100
We can conclude that 100 regular cupcakes were sold.
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°.
Answer:
No solution
Step-by-step explanation:
Absolute value can not output a negative result
