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4688-3) -30X-5 CX-5)

pshichka [43]

Answer:

x = 13/3

Step-by-step explanation:

4(8x-3) - 30x = 5(x-5)

Distribute

32x - 12 - 30x = 5x -25

Combine like terms

2x -12 = 5x-25

Subtract 2x from each side

2x-12-2x = 5x-2x -25

-12 = 3x-25

Add 25 to each side

-12+25 = 3x-25+25

13 = 3x

Divide each side by 3

13/3 = x

4 0

3 years ago

Read 2 more answers

Each cube in the prisms below has a volume of 1 cubic unit. which prism has a volume of 60 cubic units

mr_godi [17]

For A, 5x4x3=60
for B 5x3x2=30
for C 5x4x1=20
for D 5x4x2=40
hence, A has a volume of 60cubic units!

7 0

3 years ago

A group of students surveyed chose baseball and soccer as their favorite sport in the ratio 3:8

Anna [14]

$3x-\ those\ who\ chose\ baseball\\8x-\ those\ who\ chose\ soccer\\\\
a)\\3x+8x=132\\\\
11x=132\ \ \ | divide\ by\ 11\\\\x=12\\\\
8x=8*12=96\\\\
96\ chose\ soccer.\\\\\
b)\\8x=56\ \ \ | divide\ by\ 8\\\\x=7\\\\
3x=3*7=21\\\\
21\ chose\ baseball.$

5 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

Is it possible for two numbers to have a difference of 8 and a sum of 1?

topjm [15]

Answer: yes , it is possible

Step-by-step explanation:

Let the first number be x and the second number be y .Then , the sum of the two numbers will be x + y , and their difference will be x - y.

Combining the two , we have :

x + y = 1 ............................... equation 1

x - y = 8 ................................. equation 2

solving the system of linear equation by substitution method. From equation 1 , make x the subject of the formula ,

x = 1 - y ...................... equation 3

Substitute equation 3 into equation 2 ,

1 - y - y = 8

1 - 2y = 8

add 2y to both sides

1 = 8 + 2y

subtract 8 from both sides

1 - 8 = 2y

- 7 = 2y

divide through by 2

y = $\frac{-7}{2}$

y = - 3.5

substitute y = -3.5 into equation 3 to find the value of x , we have

x = 1 - y

x = 1 - ( - 3.5 )

x = 1 + 3.5

x = 4.5

Let us check :

x + y will be

4.5 + (-3.5) = 1

Also ,

x - y will be

4.5 - (-3.5)

⇒ 4.5 + 3.5 = 8

7 0

3 years ago