The teacher separated her class of twenty -eight students into two groups. One group has 4 more than twice as many students as the other group. how many students are in each group?

Answers

Answer 1

Answer: The class has 28 student. One group has four more students so surat 4 from 28. 28-4=24 since one of the groups have twice as many students as the other divide 24 by 3. 24 / 3 = 8. It says 1 group has for more than twice the other group so that would mean 2/3 of 24 and 4 would be the answer 16+4=20. The first group had 8 students, the other group has 20 students.

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Solve 8m² + 20m = 12 for m by factoring

Please show all work

Answers

$8m^(2)+20m=12 / :2$
$4m^(2)+10m=6$
$4m^(2)+10m-6 =0$
$\Delta=b^(2)-4ac$
$\Delta=100-(4*4*-6)$
$\Delta=100+96=196$
$x1=(-b-√(\Delta))/(8)$
$x1=(-10-14)/(8)$
$x1=(-24)/(8)$
$x1=-3$
$x1=(-b+√(\Delta))/(8)$
$x2=(-10+14)/(8)$
$x2=(1)/(2)$

so you can write $8m^(2)+20m=12 / :2$ as

$8(m+3)(m-(1)/(2))=8*[m^(2)+2.5m-1.5]=0$

Solve this asap thanks !

Answers

W= -56. Please mark as brainliest or like I’m trying to level up and if u need a explanation let me know

The range of the function f(k) = k2 + 2k + 1 is {25, 64}. What is the function’s domain?{5, 8}
{-5, -8}
{3, 8}
{4, 7}
{4, 8}

Answers

Answer:

Hence, the domain of the function is:

{4,7}

Step-by-step explanation:

We are given the range of the function:

$f(k)=k^2+2k+1$ as {25,16}

The function f(k) could also be written as:

$f(k)=k^2+2k+1=k^2+k+k+1\n\nf(k)=k(k+1)+1(k+1)\n\nf(k)=(k+1)(k+1)\n\nf(k)=(k+1)^2$

The range is the value of the function at some k.

if f(k)=25 then we have to find the value of k.

$f(k)=25=5^2=(k+1)^2$

on taking square root on both side we have:

$k+1=5\n\nk=5-1\n\nk=4$

if f(k)=64 then we have to find the value of k.

$f(k)=64=8^2=(k+1)^2$

on taking square root on both side we have:

$k+1=8\n\nk=8-1\n\nk=7$

Hence, the domain of the function is:

{4,7}

The domain of a function contains the x-values of the function while the range of the function contains the y-values of the function. In this case, we substitute 25 and 64 to y. Then,
25 = k2 + 2 k + 1 (k-4) * (k+6) = 0
64 = k2 + 2 k + 1 (k-7) * (k+9) = 0
hence the domain is {-9,-6, 4,7}

Use the table below to answer the questions.