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3 years ago

What is the 41st term of the sequence 8,10,12,14....

Mathematics

2 answers:

Artyom0805 [142]3 years ago

5 0

The 41th term would be 96.

Tems11 [23]3 years ago

3 0

88 i hope its help you ok

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Help me asap Please!!!

BabaBlast [244]

3.14, 2.2, 2.85, 4.4

In order they would be

Sqrt(5), Sqrt(8), pi, 2Sqrt(5)

7 0

3 years ago

Mrs. Robinson is riding the stationary bike at the gym. She completes 24 9/10 wheel rotations every 1/5 minute. What is Mrs. Rob

sweet [91]

24 9/10 x 5
124.5 rotations/minute

5 0

3 years ago

Simplify 7y + 3y - 17y.

tangare [24]

They are all like-terms so you can add 7y+3y which equals 10y. Then subtract 10y-17y and you get a negative number; -7y

5 0

3 years ago

Read 2 more answers

find all possible value of the given variable

mamaluj [8]

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

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

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

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

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

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

$7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0$

$8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0$

$9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12$

$x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0$

$10.\\ \\d^2-13d+22=0 \\ \\a=1,\ b=-13, \ q=22 \\ \\\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\\ \\ d_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{13-\sqrt{81}}{2 }=\frac{ 13-9}{2}=\frac{4}{2}=2$

$d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0$

7 0

3 years ago

A circular cake is 360 degrees if you eat a piece that is 40 degrees of the cake, how many degrees are left in the cake?

ololo11 [35]

Step-by-step explanation:

a whole circle is 360°.

if you take away 40°, how many are left ?

that is absolutely the same as the question, when I have 15 apples and 6 are eaten, how many are left ?

well, 15-6 = 9 apples are left.

and so, 360-40 = 320° are left.

7 0

2 years ago