(x-4)/3 ≤ 5<br> x ≤ ?<br> HELP ME PLEASE

One eighth of a number is ten less than one third of the number. Find the number.

GarryVolchara [31]

X/8 - 10 = x/3

bring the values with the variable on one side

x/3 - x/8 = 10

find a common denominator and subtract these fractions

8x/24 - 3x/24 = -10

now subtract fractions

8x - 3x = 5x

5x/24 = -10

multiply both sides by 24

5x = -240

divide both sides by 5

x = -48

the number is -48

7 0

3 years ago

Read 2 more answers

Sofas being sold for $603 this is a 33% discount from the m what is the original price

fgiga [73]

About $801.99 bc 603X.33 is 198.99 plus 603 is 801.99

6 0

3 years ago

JK and LM are straight lines. Find /NOK.<br> ( help me pls)<br>

VashaNatasha [74]

Answer:

$NOK = 58$

Step-by-step explanation:

Given

$JOM = 108$

$LON = 50$

Required

Find NOK

From the attached triangle, we have:

$LOK = JOM =108$ --- Corresponding angles

And

$LOK = LON + NOK$

Substitute for LOK and LON

$108= 50+ NOK$

Make NOK the subject

$NOK = 108 - 50$

$NOK = 58$

8 0

3 years ago

A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function

dexar [7]

Here's our equation.

$h=-16t+64t+3$

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

$0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}$

$= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}$

So the ball will return to the ground at the positive value of $\boxed{\frac{8+\sqrt{67}}{4}}$ seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

$\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2$

$h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}$

8 0

4 years ago

A triangular pyramid has a base with an area of 43 cm squared and lateral faces with bases of 10 cm and heights of 8.6 cm. What

Jet001 [13]

Answer:

The total surface area of triangular pyramid is 172 cm squared

Step-by-step explanation:

Triangular pyramid:

Number of faces 4.
Number of vertices of a triangular pyramid is 6.
The volume is $\frac13 AH$ . A= area of the pyramid's base and H= height of the pyramid.