What are three possible solutions for X>10/8

4. Kristina is allowed to have a friend sleep over every 3rd weekend,

PSYCHO15rus [73]

Answer:

6

Step-by-step explanation:

You must find the least common multiple of their schedules.

The numbers 3 and 2 are their own prime factors, so the least common multiple of 3 and 2 is 6.

Six weekends must pass until they can both have someone over on the same night.

The number line below shows that the first time their sleepovers coincide is Week 6.

3 0

2 years ago

The area of a piece of land that is in the shape of a triangle is 1/12 square mile. One dimension of this piec of land is 4/21.

Arisa [49]

Answer:

D2=21/24 mile

Step-by-step explanation:

Her it is not specified if it is base or height, it only talks about dimensions, so let's work with dimensions

A=1/12 $mile^{2}$

D1=4/21mile, D2=?

Area of a triangle $A=\frac{b.h}{2}$

A=((D1*D2)/2) → 2A=D1*D2 → D2=2A/D1 so D2=(2*1/12)/(4/21) =(1/6)/(4/21) = (1*21)/(6*4)=21/24 mile

7 0

3 years ago

John’s teacher gave him the following system of equations and asked him to find the point of intersection. 3x + 2 = y 2x – 3= y

Galina-37 [17]

1) to this case you must match both equations

3x+2 = 2x-3 ⇒3x-2x = -3-2⇒x= -5

now the value of x substitute it in the any of two equations, y = 3(-5)+2 = -13

and the other equation is the same value

y = 2(-5)-3 = -13

the point is ( -5,-13)

7 0

3 years ago

What does 9/5 equals to ?

goldenfox [79]

Answer:

9/5 = 1.8

All you have to do is divide the numerator by the denominator.

Hope this helps!

-Mikayla

8 0

3 years ago

Read 2 more answers

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: Rate of change measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

2 years ago