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3 years ago

8x= 3x + 22 What's the solution?

2 answers:

8 0

Step 1: Subtract 3x from both sides.8x−3x=3x+22−3x5x=22Step 2: Divide both sides by 5.5x5=225x=225Answer:x=225

Diano4ka-milaya [45]3 years ago

5 0

Well you subtract 3 from both sides then you get 5x and then divide it by 22 and then your answer is a decimal 4.4

You might be interested in

Arithmetic sequences et sn} be an arithmetic sequence that starts with an initial index of 0. The initial term is 3 and the comm

navik [9.2K]

Answer:

(a) The value of $s_z$ is (z+1)(3-z).

(b) The next term in the sequence is -2.

Step-by-step explanation:

(a)

It is given that arithmetic sequence that starts with an initial index of 0.

The initial term is 3 and the common difference is -2.

$a_0=3$

$d=-2$

We need to find the value of $s_z$ .

$s_z=\sum_{n=0}^{n=z}(a+nd)$

where, a is initial term and d is common difference.

$s_z=\sum_{n=0}^{n=z}(3-2n)$

The sum of an arithmetic sequence with initial index 0 is

$s_n=\frac{n+1}{2}[2a+nd]$

where, a is initial term and d is common difference.

Substitute n=z, a=3 and d=-2 in the above formula.

$s_z=\frac{z+1}{2}[2(3)+z(-2)]$

$s_z=\frac{z+1}{2}[2(3-z)]$

$s_z=(z+1)(3-z)$

Therefore the value of $s_z$ is (z+1)(3-z).

(b)

The given arithmetic sequence is

7, 4, 1, ...

We need to find the term in the sequence.

In the given arithmetic sequence the first term is

$a=7$

The common difference of the sequence is

$d=a_2-a_1\Rightarrow 4-7=-3$

The first term is 7 and common difference is -3.

Add common difference in last given term, i.e., 1, to find the next term of the sequence.

$1+(-3)=1-3=-2$

Therefore the next term in the sequence is -2.

3 0

4 years ago

The vertices of a square CDEF are C(1,1), D(3,1), E(3,-1) and F(1,-1). What formulas prove that the diagonals are congruent perp

Oxana [17]

To prove that the diagonals are congruent, you need to formula to compute the distance between two points:

$d(A,B) = \sqrt{(A_x-B_x)^2 + (A_y-B_y)^2}$

Using that formula, you may prove that $d(C,E) = d(D,F)$ , which means that the two diagonals have the same length.

To prove that they are perpendicular, you need the formula to compute the slope of a segment. The slope, knowing the enpoints, is given by

$m = \cfrac{\Delta y}{\Delta x} = \cfrac{A_y-B_y}{A_x-B_x}$

You can use this formula to prove that

$m_{CE} = -\cfrac{1}{m_{DF}}$

In fact, if one slope is the opposite of the reciprocal of the other, the two segments are perpendicular.

Finally, to prove that they bisect each other, you first need to find the point where they meet. First of all, you need to find the line the segments lie on: the formula is

$y-y_0 = m(x-x_0)$

where $(x_0,y_0)$ is one of the points belonging to the line, and you already know how to find the slope. Then, you find the point of intersection, say A, by solving the system involving the two lines:

$\begin{cases} y= m_{CE}x+q_{CE}\\ y = m_{DF}x+q_{DF} \end{cases}$

And use again the formula for the distance between two points to prove that

$d(A,C) = d(A,D) = d(A,E) = d(A,F)$

4 0

3 years ago

Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:

Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0

3 years ago

Five fourths of a right angle

sleet_krkn [62]

The answer would be 112.5 plz give me thanks and full rating PLZZZZZZ

8 0

3 years ago

Please answer now ! 90 points

ozzi

Step-by-step explanation:

2+3b $\leq$ 25

b=books

3b $\leq$ 23

$b\leq 23/3$

$b\leq 7.67$

50+25L $\leq$ 200

L=lesson

25L $\leq$ 150

L $\leq$ 6

Hope that helps :)

7 0

3 years ago

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