Let's say our first integer is "a".
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
![\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20times%20the%20first%7D%7D%7B3a%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B26%20less%20than%20twice%20the%20sum%20of%20the%20others%7D%7D%7B2%5B~%28a%2B2%29%2B%28a%2B4%29~%5D~~~-26%7D%0A%5C%5C%5C%5C%5C%5C%0A3a%3D2%5B~2a%2B6~%5D-26%5Cimplies%203a%3D4a%2B12-26%5Cimplies%203a%3D4a-14%0A%5C%5C%5C%5C%5C%5C%0A0%3Da-14%5Cimplies%2014%3Da)
what are the other two consecutive integers? well, a + 2 and a + 4.
#1)
A) b = 10.57
B) a = 22.66; the different methods are shown below.
#2)
A) Let a = the side opposite the 15° angle; a = 1.35.
Let B = the angle opposite the side marked 4; m∠B = 50.07°.
Let C = the angle opposite the side marked 3; m∠C = 114.93°.
B) b = 10.77
m∠A = 83°
a = 15.11
Explanation
#1)
A) We know that the sine ratio is opposite/hypotenuse. The side opposite the 25° angle is b, and the hypotenuse is 25:
sin 25 = b/25
Multiply both sides by 25:
25*sin 25 = (b/25)*25
25*sin 25 = b
10.57 = b
B) The first way we can find a is using the Pythagorean theorem. In Part A above, we found the length of b, the other leg of the triangle, and we know the measure of the hypotenuse:
a²+(10.57)² = 25²
a²+111.7249 = 625
Subtract 111.7249 from both sides:
a²+111.7249 - 111.7249 = 625 - 111.7249
a² = 513.2751
Take the square root of both sides:
√a² = √513.2751
a = 22.66
The second way is using the cosine ratio, adjacent/hypotenuse. Side a is adjacent to the 25° angle, and the hypotenuse is 25:
cos 25 = a/25
Multiply both sides by 25:
25*cos 25 = (a/25)*25
25*cos 25 = a
22.66 = a
The third way is using the other angle. First, find the measure of angle A by subtracting the other two angles from 180:
m∠A = 180-(90+25) = 180-115 = 65°
Side a is opposite ∠A; opposite/hypotenuse is the sine ratio:
a/25 = sin 65
Multiply both sides by 25:
(a/25)*25 = 25*sin 65
a = 25*sin 65
a = 22.66
#2)
A) Let side a be the one across from the 15° angle. This would make the 15° angle ∠A. We will define b as the side marked 4 and c as the side marked 3. We will use the law of cosines:
a² = b²+c²-2bc cos A
a² = 4²+3²-2(4)(3)cos 15
a² = 16+9-24cos 15
a² = 25-24cos 15
a² = 1.82
Take the square root of both sides:
√a² = √1.82
a = 1.35
Use the law of sines to find m∠B:
sin A/a = sin B/b
sin 15/1.35 = sin B/4
Cross multiply:
4*sin 15 = 1.35*sin B
Divide both sides by 1.35:
(4*sin 15)/1.35 = (1.35*sin B)/1.35
(4*sin 15)/1.35 = sin B
Take the inverse sine of both sides:
sin⁻¹((4*sin 15)/1.35) = sin⁻¹(sin B)
50.07 = B
Subtract both known angles from 180 to find m∠C:
180-(15+50.07) = 180-65.07 = 114.93°
B) Use the law of sines to find side b:
sin C/c = sin B/b
sin 52/12 = sin 45/b
Cross multiply:
b*sin 52 = 12*sin 45
Divide both sides by sin 52:
(b*sin 52)/(sin 52) = (12*sin 45)/(sin 52)
b = 10.77
Find m∠A by subtracting both known angles from 180:
180-(52+45) = 180-97 = 83°
Use the law of sines to find side a:
sin C/c = sin A/a
sin 52/12 = sin 83/a
Cross multiply:
a*sin 52 = 12*sin 83
Divide both sides by sin 52:
(a*sin 52)/(sin 52) = (12*sin 83)/(sin 52)
a = 15.11
Answer:
the numbers are -20 and 5
Step-by-step explanation:
I think the answer is Beth can make 8 dozen gingerbread men
Number of weekend minutes used: x
Number of weekday minutes used: y
This month Nick was billed for 643 minutes:
(1) x+y=643
The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44
We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44
Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y
Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44
Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72
Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243
Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400
Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
