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3 years ago

How many Solutions does 6 + x/5 = x/5 - 5 - 3x/10 have?

Mathematics

1 answer:

klasskru [66]3 years ago

6 0

9514 1404 393

Answer:

1 solution (x = -36 2/3)

Step-by-step explanation:

6 + x/5 = x/5 - 5 - 3x/10 . . . . . . . given

60 +2x = 2x -50 -3x . . . . . . . . . multiply by 10

3x = -110 . . . . . . . . . . . . . . . . add x-60 to both sides

x = -110/3 = -36 2/3 . . . . . .divide by the coefficient of x

There is one solution to the equation.

_____

Additional comment

The coefficient of x on the left side of the equation is not equal to the coefficient of x on the right side. When that is the case, the equation will have one solution.

When the coefficients are equal, there will be no solutions if the constants are different, and there will be infinite solutions if the constants are the same.

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C = 0.20m + 2.00 what is the charge for 1mile ride?

RUDIKE [14]

All you have to do is substitute the 1 into the m for the mile ride:
C = 0.20(1) + 2.00 = 0.20 + 2.00 = 2.20

The charge for 1 mile ride is $2.20.

5 0

4 years ago

Consider the experiment where a pair of fair dice is thrown. Let X denote the random variable whose value is determined by takin

m_a_m_a [10]

Answer:

P(X=1) = 1/36

P(X=2) = 3/36 = 1/12

P(X=3) = 5/36

P(X=4) = 7/36

P(X=5) = 9/36 = ¼

P(X=6) = 11/36

7/36

Step-by-step explanation:

Attached, you will find a table with the possible values of X.

(See table attached)

As we can see, there are 36 entries so,

P(X=1) = 1/36

P(X=2) = 3/36 = 1/12

P(X=3) = 5/36

P(X=4) = 7/36

P(X=5) = 9/36 = ¼

P(X=6) = 11/36

It can be noticed that the only possible value of X which is divisible by 4 is 4 and there are 7 entries in the table, so

P(X divisible by 4) = P(X=4) = 7/36

3 0

3 years ago

Complete the set of ordered pairs for the relation.

Nonamiya [84]

The set of ordered pairs are $(-2,18),(-1,20),(0,22),(1,24),(2,26)$

Step-by-step explanation:

Given that the relation $\{(x, y): y=2(x+11) \text { and } x \in(-2,-1,0,1,2)\}$

To find the set of ordered pairs, let us substitute the value for x in the relation $y=2(x+11)$

For $x=-2$ ,

$\begin{aligned}y &=2(-2+11) \\&=2(9) \\y &=18\end{aligned}$

The ordered pair is $(-2,18)$

For $x=-1$ ,

$\begin{aligned}y &=2(-1+11) \\&=2(10) \\y &=20\end{aligned}$

The ordered pair is $(-1,20)$

For $x=0$ ,

$\begin{aligned}y &=2(0+11) \\&=2(11) \\y &=22\end{aligned}$

The ordered pair is $(0,22)$

For $x=1$ ,

$\begin{aligned}y &=2(1+11) \\&=2(12) \\y &=24\end{aligned}$

The ordered pair is $(1,24)$

For $x=2$ ,

$\begin{aligned}y &=2(2+11) \\&=2(13) \\y &=26\end{aligned}$

The ordered pair is $(2,26)$

Thus, the set of ordered pairs is $(-2,18),(-1,20),(0,22),(1,24),(2,26)$

7 0

3 years ago

Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM

denis23 [38]

Answer:

The length of side KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid)

$EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of side KM is $\sqrt{109}$ units.

8 0

3 years ago

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What do you know about a quadratic equation that has the following points (2, 0) and (10, 0)?

Agata [3.3K]

Answer:

C. The x-coordinate of the vertex must be 6

Step-by-step explanation:

The parabola intercepts the x-axis when y = 0.

Therefore, if the quadratic equation has the points (2, 0) and (10, 0) then the x-intercepts or "zeros" are x = 2 and x = 10.

The x-coordinate of the vertex is the midpoint of the zeros.

$\sf \implies midpoint=\dfrac{2+10}{2}=6$

Therefore, the solution is option C.

Additional Information

The leading coefficient of a quadratic tells us if the parabola opens upwards or downwards:

Positive leading coefficient = parabola opens upwards
Negative leading coefficient = parabola opens downwards

We have not been given this information and so therefore cannot determine the way in which it opens.

As we do not know the way in which way the parabola opens, we cannot determine if the parabola will have a negative or positive y-intercept.

We have not been given the full quadratic equation, and so we cannot determine if the parabola is wider (or narrower) than the parent function.

3 0

3 years ago

Read 2 more answers