Eight more than the quotient of a number and 4 equals 5

1 answer:

3 0

Well we can put that word problem into a equation so, 8+x/4=5. Now what I would do is move the 8 to the other side, and since it moved to the other side it becomes a -8 so now we have x/4 = -3. Now I would multiply both side by 4 to get x = -12

Answer= X = -12

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Can anybody help me plzz

Fittoniya [83]

Answer:

1 : 12 : 4

Step-by-step explanation:

Let A, G, C represent the ages of Alex, George, and Carl, respectively. We are told that ...

A = G +12

C = 3A . . . . . . . we read this as "3 times as old"; not "3 times older"

A + G + C = 68

Substituting for G and C, we have ...

A + (A -12) +(3A) = 68

5A = 80 . . . . . add 12

A = 16 . . . . . . . divide by 5

The other ages are then ...

G = A-12 = 16 -12 = 4

C = 3A = 3·16 = 48

The ratios are ...

G : C : A = 4 : 48 : 16

G : C : A = 1 : 12 : 4

7 0

3 years ago

Find the distance of the line segment joining the two points (-4 /2 - /12) and (/32, 2/3)

astraxan [27]

Answer: $4\sqrt{3}$ .

Step-by-step explanation:

Distance formula : Distance between points (a,b) and (c,d) is given by :-

$D=\sqrt{(d-b)^2+(b-a)^2}$

Distance between points $(-4\sqrt{2},\sqrt{12}) \text{ and }(-\sqrt{32}, 2\sqrt{3})$ .

$D=\sqrt{(2\sqrt{3}-(-\sqrt{12}))^2+(-\sqrt{32}-(-4\sqrt{2}))}\\\\=\sqrt{(2\sqrt{3}+\sqrt{2\times2\times3})^2+(-\sqrt{4\times4\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}-\sqrt{2^2\times3})^2+(-\sqrt{4^2\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}+2\sqrt{3})^2+(-4\sqrt{2}+4\sqrt{2})^2}\\\\=\sqrt{(4\sqrt{3})^2+0}\\\\=4\sqrt{3}\text{ units}$

Hence, the correct option is $4\sqrt{3}$ .

3 0

3 years ago

Evaluate the determinant for the following matrix: 1 4 4 5 2 2 1 5 5

Elina [12.6K]

A Laplace expansion along the first row gives

$\begin{vmatrix}1&4&4\\5&2&1\\1&5&5\end{vmatrix}=1\begin{vmatrix}2&1\\5&5\end{vmatrix}-4\begin{vmatrix}5&1\\1&5\end{vmatrix}+4\begin{vmatrix}5&2\\1&5\end{vmatrix}$
$=1(10-5)-4(25-1)+4(25-2)$
$=1$