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3 years ago

The following data gives the number of cars

Mathematics

1 answer:

faltersainse [42]3 years ago

5 0

Answer:

Step-by-step explanation:

2 : (2 + 9 + 16 + 12 + 8 + 6 + 4 + 2 + 1) = 1/30 ≈ 3.33% ≈ 3%

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Martina is currently 14 years older than her cousin Joey. In 5 years she will be 3 times as old as Joey. Use this information to

Komok [63]

<span>1. Martina is given to be 14 years older than Joey who is x years old. Hence, the age of Martina can be expressed as, x + 14
2. If Joey's current age is x then in 5 years he will become x+5 years old.
3. Martina's age will be her current age added with 5 making it, x+14+5 or x + 19.
4. The equation that would best represent the given scenario in this item is: x + 19 = 3(x + 5)
5. From the given above, the value of x is equal to 2. This is Joey's current age.
6. The current age of Martina is x + 14 which is equal to 16.</span>

3 0

3 years ago

4 ∈ {4, 14, 17} is this true or false

Natalija [7]

The symbol "<span>∈" means "belongs to".
This means that a certain element belongs to or is found in the set.

The set given contains three numbers which are 4, 14 and 17.
As you can see, 4 is found in this set, therefore, 4 belongs to this set.

Based on this, the given statement is true.</span>

5 0

3 years ago

3x+6<25,x=8 true or false

Liono4ka [1.6K]

That is false

If x=8 then the whole thing is 24+6 which is 30

so 30<25 is false

6 0

3 years ago

Read 2 more answers

Jack has two more than half as many cups as

Rufina [12.5K]

Step-by-step explanation:

$(x \div 2 + 2) + \frac{(x \div 2 + 2)}{2} + x$

its hard to determine without the known value of x, but the above would be the equation needed to solve it if you knew what x was

4 0

3 years ago

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

3 years ago