All categories

2 years ago

Kevin is 6 years older than Timothy. 4 years ago, Kevin was twice as old as Timothy. Find their present ages.

Mathematics

1 answer:

Damm [24]2 years ago

3 0

Answer:

Kevin is 16; Timothy is 10

Step-by-step explanation:

k= Kevin's age

t=Tim's age

k=t+6

k-4=2(t-4)

k-4=2t+8

(t+6)-4=2t-8 :substitution

t+2=2t-8

t=10

k=t+6

k=10+6

k=16

You might be interested in

Find the product of (3.7 × 104) and 2.

Lera25 [3.4K]

<span>product of (3.7 × 104) and 2
</span>
answer is A.) 7.4 × 104

7 0

3 years ago

Read 2 more answers

Find the equation of the line passing through (-2,-9) and (-1,-1).

leonid [27]

Answer:

y=8x+25

Step-by-step explanation:

find the slope of the points

put it in point slope form

than distribute

than add or subtract

and you find your answer of y=mx+b

7 0

3 years ago

The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is

bulgar [2K]

The integral is approximated by the sum,

$\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))$

where $f(x)=x^3$ and $x_n=\dfrac12n$ , giving you

$\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)$
$\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)$
$\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)$

Faulhaber's formulas make short work of computing the sum. You have

$\displaystyle\sum_{n=0}^k1=k+1$
$\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2$
$\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6$
$\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4$

which gives

$\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31$
$\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}$
$\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25$

4 0

3 years ago

Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to the

Georgia [21]

Answer: The answers are

(i) The slope of segments DE and AC is not 0.

(ii) The coordinates of D and E were found using the Midpoint Formula.

Step-by-step explanation: We can easily see in the proof that the co-ordinates of D and E were found using the mid-point formula, not distance between two points formula. So, this is the first flaw in the Gina's proof.

Also, we see that the slope of line DE and AC, both are same, not equal to 0 but is equal to

$\dfrac{y_2}{x_2},$