Cu cat este egal 5! ?

Solve the inequality –3v – 2v > 35 for v.

viva [34]

Answer:

-7 > v OR vice-versa [v < -7]

Step-by-step explanation:

Combine like-terms [-5v > 35], then divide by -5.

NOTE: Since you are dividing by a negative, reverse the sign.

3 0

3 years ago

Read 2 more answers

Solve the following equations using substitution.

gayaneshka [121]

Answer:

x = 3.75

y= 0.5

Step-by-step explanation:

2x and 3y and 6

2x and y and 7

the difference between those two is 2y and 1

because 3y-y = 2y and 7-6 = 1

2y=1 y = 0.5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2x - 3 = 6 is also 6 + 3y = 2x

3(0.5) = 1.5

6 + 1.5 = 75.5

2x = 7.5

x= 3.75

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

check:

2(3.75) = 7.5

3(0.5) = 1.5

7.5-1.5 = 6

2(3.75) = 7.5

y= (0.5)

7.5-0.5= 7

3 0

3 years ago

If f(x) varies directly with x^2, and f(x) = 75 when x = 5, find the value of f(4). A)100 B)50 C)48 D)25

ehidna [41]

That means f(x)=kx²

so
f(5)=75=5²k
solve for k
75=5²k
75=25k
divide by 25
3=k

f(x)=3x² is da equation

so
f(4)=3(4)²
f(4)=3(16)
f(4)=48

C is answer

6 0

4 years ago

A million years ago, an alien species built a vertical tower on a horizontal plane. When they returned they discovered that the

velikii [3]

Answer:

$\theta=cos^{-1}(\frac{2}{3})$

Step-by-step explanation:

We are given that measurements of three points on the ground gave coordinates of (0,0,0),(1,2,0) and (0,2,1)

We have to find the angle by which the tower now deviate from the vertical

We find cross product of <1,2,0> and <0,2,1>

$\times =\begin{vmatrix}i&j&k\\1&2&0\\0&2&1\end{vmatrix}$

$\times =2\hat{i}-\hat{j}+2\hat{k}$

Now, we are finding the angle between $\times$ and vertical vector <0,0,1>

Angle between two vectors formula

$cos\theta=\frac{a.b}{\mid a\mid\cdot\mid b\mid }$

Now, using this formula

$cos\theta=\frac{2}{1\cdot 3}=\frac{2}{3}$

$\theta=cos^{-1}(\frac{2}{3})$

Hence, the tower deviate from the vertical by the angle $\theta=cos^{-1}(\frac{2}{3})$

6 0

3 years ago

X^3>x^2 and |x| > 1, then

Citrus2011 [14]

Answer:

1. X > 1

Step-by-step explanation:

We know $x^2 > 1$ because $x^2 = |x|^2 > 1$ . That's why

$X^3 > x^2 > 1$ .

We can now subtract 1 in both sides of the inequality:

$X^3 -1 > 0$ .

Factoring $X^3 -1$ as a difference of cubes, we get:

$(X-1)(X^2 + X + 1) > 0$ .

Thus, we have two factors whose multiplication is positive. Then, both are positive or both are negative. The second case is impossible, because $X^2 + X + 1$ can never be positive. The reason is the following:

$X^2 + X + 1 = 2 \frac{X^2 + X + 1}{2} = \frac{2X^2 + 2X + 2}{2} = \frac{X^2 + 2X + 1 +X^2 + 1}{2} = \frac{(X+1)^2 +X^2 + 1}{2}$

witch is always positive because $(X+1)^2 + X^2 + 1$ is a sum of squares, and the squares are always positive.

We conclude that both $(X-1)$ and $(X^2 + X + 1)$ have to be positive, and then $X-1 > 0$ implies $X > 1$

4 0

3 years ago