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1 year ago

14 is 10% of ____? Or what is 14 10% of

Mathematics

1 answer:

leonid [27]1 year ago

3 0

Answer:

THE ANSWER TO THE TOP LINE IS: 140

Step-by-step explanation:

why? becuase 10 times 10 is 100 all whole fractions need to be ?/100 (something over 100) so then becuase you did 10 times ten to find what 14 is you do 14 times 10. it should look like this :D.

10*10=100

14*10=140 . 140/100 is your answer or 140 :)

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Find measure of Arc BC

iren2701 [21]

Answer:

arc BC = 140

Step-by-step explanation:

Comment

Pick a point inside the circle. It should look like this point is the center of the circle. Call this point D. Draw lines CD and BD

CD makes a right angle with AC

BD makes a right angle with AB

The central angle formed by The given information above is

<BDC = 360 - 90 - 90 - 40

<BDC = 140.

Answer The arc measurement of BC = 140

5 0

1 year ago

What is the recursive rule for the following sequence: -5, -3, -1, 1…

Irina18 [472]

$a_{n}$ = $a_{n-1}$ + 2

So C.

4 0

2 years ago

If g(x) = -4x + 5, find g(2x-1).

lilavasa [31]

Answer:

- 8x + 9

Step-by-step explanation:

To evaluate g(2x - 1) substitute x = 2x - 1 into g(x), that is

g(2x - 1)

= - 4(2x - 1) + 5 ← distribute parenthesis and simplify

= - 8x + 4 + 5

= - 8x + 9

8 0

3 years ago

Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Given:

lisov135 [29]

Answer:

Step-by-step explanation:

6 0

2 years ago