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3 years ago

(Help)Find the value of X. 15-18

Mathematics

1 answer:

Juliette [100K]3 years ago

6 0

Answer:Tag 30*(2\/3)=x

Step-by-step explanation:

Tag 30*= x/2\/3;

Tag 30*(2\/3)=x

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Simplify: 14 + 8 - 20:5+ 2(4 - 1)

ki77a [65]

Answer:

2:5

Step-by-step explanation:

14 + 8 = 22 - 20 = 2

4-1 = 3 x 2 = 5 + 5 = 10

2:5

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3 years ago

Write an algebraic rule to describe the translation B(3,2)→ 8'(-5,0).

kompoz [17]

Answer:

<-8,-2>

Step-by-step explanation:

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3 years ago

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Factor 3x^3+3x^2+x+1

Nat2105 [25]

We can solve it like:
3x^3+3x^2+x+1= (3x^3+3x^2)+(x+1)=
3x^2(x+1)+(x+1)=
= (x+1)(3x^2+1)

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4 years ago

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50 points, question is below.

REY [17]

$x^2+3x-10\neq0\\\\x^2+5x-2x-10neq0\\\\x(x-5)-2(x+5)\neq0\\\\(x+5)(x-2)\neq0\to x+5\neq0\ \wedge\ x-2\neq0\\\\x\neq-5\ \wedge\ x\neq2\\\\\dfrac{x+5}{x^2+3x-10}=\dfrac{x+5}{(x+5)(x-2)}=\dfrac{1}{x-2}\\\\Answer:\ x=2$

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4 years ago

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The velocity of a turtle is recorded at 1 minute intervals (in meters per second). Use the right-endpoint approximation to estim

aliina [53]

Answer:

The total distance that the turtle traveled during the 5 seconds recorded is $Distance \:traveled\approx 3.838\:\frac{m}{s}$

Step-by-step explanation:

To estimate distance traveled of an object moving in a straight line over a period of time, from discrete data on the velocity of the object, we use a Riemann Sum. If we have a table of values

$\left\begin{array}{ccccccc}time\:=\:t_i&t_0=0&t_1&t_2&...&t_n\\velocity\:=\:v(t_i)&v(t_0)&v(t_1)&v(t_2)&...&v(t_n)\end{array}\right$

where $\Delta t=t_i-t_{i-1}$ , then we can approximate the displacement on the interval $[t_{i-1},t_i]$ by $v(t_{i}) \times\Delta t$ .

Therefore the distance traveled of the object over the time interval $[0,t_n]$ can be approximated by

$Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+...+|v(t_n)|\Delta t$

This is the right endpoint approximation.

We are given a table of values for <em>v(t)</em>

$\left\begin{array}{cccccccc}t(sec)&0&1&2&3&4&5\\v(t)&0.078&0.83&0.75&0.98&0.853&0.425\end{array}\right$

Applying the right endpoint approximation formula we get,

$\Delta t = 1\sec$

$Distance \:traveled\approx |v(t_1)|\Delta t+|v(t_2)|\Delta t+|v(t_3)|\Delta t+|v(t_4)|\Delta t+|v(t_5)|\Delta t\\\\Distance \:traveled\approx 0.83(1)+0.75(1)+0.98(1)+0.853(1)+0.425(1)\\\\Distance \:traveled\approx 3.838\:\frac{m}{s}$

The total distance that the turtle traveled during the 5 seconds recorded is $Distance \:traveled\approx 3.838\:\frac{m}{s}$

4 0

3 years ago