1.) 8/1 - 1 and 11/16
2.) 1 and 3/5 + 1 and 1/4
3.) 5 and 1/4 - 1 and 1/8

Answers

Answer 1

Answer: $^(16))8-1(11)/(16)= (128)/(16)- (1\cdot16+11)/(16)= (128-27)/(16)=\bold{ (101)/(16)} \n \n 1 (3)/(5)+1 (1)/(4)=^(4))(1\cdot5+3)/(5) +^(5))(1\cdot4+1)/(4)= (32)/(20)+ (25)/(20)=\bold{ (57)/(20)} \n \n 5 (1)/(4)-1 (1)/(8)=^(4))(5\cdot4+1)/(4) -^(2))(1\cdot8+1)/(8)= (84)/(16)- (18)/(16)= (66)/(16)^((2)=\bold{ (33)/(8) }$

Answer 2

Answer: $\frac{8}1-1(11)/(16)=(8-1)-(11)/(16)=6(16-11)/(16)=\boxed{6(5)/(16)}$

$1\frac{3}5+1\frac{1}4=(1+1)+\frac{3}5+\frac{1}4=2+(3\cdot4+5)/(4\cdot5)=\boxed{2(17)/(20)}$

$5\frac{1}4-1\frac{1}8=(5-1)+(\frac{1}4-\frac{1}8)=4+(2-1)/(8)=\boxed{4\frac{1}8}$

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Given: PKHE - inscribed, m∠E=120°, m∠EPK=51°, PF ∥ EH .
Find: m∠KPF, m∠H.

Answers

Answer:

C is the answer they are equal.

Step-by-step explanation:

The paper cup would not roll straight. One end was 7 cm In a diameter, and the other end was 5cm in diameter. In one roll of the cup, A. How far would the larger end roll?
B. how far would the smaller end roll?
(Round each answer to the nearest centimeter. Use 3.14 for pi)

Answers

You need to find Circumference of each circle so the top is 7cm and the bottom is 5cm
C = pi • diameter
So top is 3.14 • 7 = 21.98
And bottom is 3.14 • 5 = 15.70
Need any more help?

1 half minus five twelfths

Answers

1/2 - 5/12= 6/12 - 5/12 = 1/12

very easy fined a denominator most commen and subtract

What type of triangle has 3 sides that are the same length?

Answers

an equilateral triangle.

The answer is an equilateral triangle.

An equilateral triangle is a triangle in which all 3 side lengths are the same.

Hope this helped☺☺

Find all possible value of the given variable 1.h²+5h=0
2.z²-z=0
3.m²+13m+40=0
4.z²-3z=0
5.q²+7q=0
6.k²+2k=0
7.x²-3x-70=0
8.q²+7q-60=0
9.z²+9z-36=0
10.d²-13d+22=0

Answers

$1.\n \n h^2+5h=0 \n \nh(x+5)=0\n \nx=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\n \nx+5-5=0-5\n \nx=0 \ \ \ or \ \ \ x=-5$

$2.\n \n z^2-z=0\n \nz(x-1)=0\n \nz=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\n \nz-1+1 =0 +1 \n \nx=0 \ \ \ or \ \ \ z=1$

$3.\n \nm^2+13m+40=0 \n \na=1 ,\ b=13, \ c=40 \n \n\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \n \nand \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution$

$4.\n \nz^2-3z=0 \n \n (z-3)=0\n \nz=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\n \n z-3+3=0+3\n \nz=0 \ \ \ or \ \ \ z=3$

$5.\n \nq^2+7q=0 \n \nq(q+7)=0\n \nq=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\n \nq+7-7=0-7\n \nq=0 \ \ \ or \ \ \ q=-7$

$6.\n \nk^2+2k=0\n \nk(k+2)=0\n \nk=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\n \nk+2-2=0-2\n \nk=0 \ \ \ or \ \ \ k=-2$

$7. \n \n x^2-3x-70=0 \n \na=1,\ b=-3, \ c=-70 \n \n\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\n \n x_(1)=(-b-√(\Delta) )/(2a)=(3-√(289))/(2 )=( 3-17)/(2)=(-14)/(2)=-7$

$x_(2)=(-b+√(\Delta) )/(2a)=(3+√(289))/(2 )=( 3+17)/(2)=(20)/(2)=10\n \n(x+7)(x-10)=0$

$8.\n \nq^2+7q-60=0 \n \na=1,\ b=7, \ q=-60 \n \n\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \n \n x_(1)=(-b-√(\Delta) )/(2a)=(-7-√(289))/(2 )=( -7-17)/(2)=(-24)/(2)=-12$

$x_(2)=(-b+√(\Delta) )/(2a)=(-7+√(289))/(2 )=( -7+17)/(2)=( 10)/(2)= 5\n \n(x+12)(x-5)=0$

$9.\n \nz^2+9z-36=0 \n \na=1,\ b=9, \ q=-36 \n \n\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\n \n x_(1)=(-b-√(\Delta) )/(2a)=(-9-√(225))/(2 )=( -9-15)/(2)=(-24)/(2)=-12$

$x_(2)=(-b+√(\Delta) )/(2a)=(-9+√(225))/(2 )=( -9+15)/(2)=(6)/(2)=3\n \n(x+11)(x-3)=0$

$10.\n \nd^2-13d+22=0 \n \na=1,\ b=-13, \ q=22 \n \n\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\n \n d_(1)=(-b-√(\Delta) )/(2a)=(13-√(81))/(2 )=( 13-9)/(2)=(4)/(2)=2$

$d_(2)=(-b+√(\Delta) )/(2a)=(13+√(81))/(2 )=( 13+9)/(2)=(22)/(2)=11\n \n(d-2)(d-11)=0$

For every $5 Micah saves, his parents give him $10.If Micah saves $150, how much money will his parents give him?