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vazorg [7]
4 years ago
6

Solve the following equations for ????. Write your answer in set notation.

Mathematics
1 answer:
ddd [48]4 years ago
7 0

Answer:

a. x=7

b. x=4

c. x=5

d. x=4

e. equation b and d have same solution set.

Step-by-step explanation:

a.

  3x=21

   x=21/3

   x=7.

b.

3x+9-5=16

3x+4=16

3x=12

x=12/3

x=4

c.

6x-9-5=16

6x-14=16

6x=30

x=30/6

x=5

d.

6x+18-10=32

6x+8=32

6x=32-8

6x=24

x=24/6

x=4

e. Equation b and d have the same solution set.

The expression on both sides of the equal in the equation d is are twice of the expression in equation b. So, if (right side)=(left side) is true for some values of x, then 2(right side) = 2(left side) will be true for the same values of x.So we can find the solution without finding the solution set of both the equations.

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Answer:

1and2and3and4and5and5

6 0
3 years ago
Task 2
Softa [21]

Answer:

A) Carla: y = 1.5x + 30   <-- y is the total money, and x is the number of weeks

Darla: y = 3x   <-- y is the total money, and x is the number of weeks

B) No, they start with a different amount of money and also save different amounts each week. They will not always have the same amount of money.

C) See attachment

Step-by-step explanation:

We'll start with Carla.

Carla saves $1.5 every week. That means that if we multiply the number of weeks she has saved by 1.5 we'll get the amount of money she has. However, she starts at 30 so we have to add 30 to that at the beginning. So, Carla's equation is y = 1.5x + 30. y is the total money, and x is the number of weeks

Darla saves $3 every week. That means that if we multiply the number of weeks she has saved by 3 we'll get the amount of money she has. Unlike Carla, Darla starts at $0, so there is no need to add anything. So, Darla's equation is y = 3x. y is the total money, and x is the number of weeks

Since Carla and Darla start with a different amount of money and save different amounts each week, they won't always have the same amount of money. However, they will cross at some point. You can actually see this in the graph. Carla, there is red, and Darla is blue.

Now let's graph the 2 equations. First, we'll do Carla. her equation is y = 1.5x + 30. Since it is in the slope-intercept form we can just use y = mx + b to tell how we're going to graph this. In this formula, m is the slope, and b is the y-intercept. As you can see when you look back at Carla's equation, m is 1.5 and b is 30. First, we'll take the y-intercept, b, which is 30 as we just figured out, and plot it on the graph. All you have to do here is count 30 units up the y-axis (the one going up and down) and put a point there. Now we take the slope (1.5) and we start putting in points that the line will go through. 1.5 in fraction form is 3/2 meaning (starting from that point we marked earlier) we will go 3 units up, and 2 units to the right and put another point there. You can now take a ruler, line up the 2 points you marked and draw a big line through them that keeps going on forever on both sides (just put an arrow at both ends to mark that).

Now we'll do the same with Darla's equation. Her's is y = 3x. When we compare this with or slope-intercept formula, y = mx + b, we'll notice there is no b. That means it is 0 and that the y-intercept is also 0. So go ahead and mark a point in the middle of the graph where the 2 big lines cross. 3 in fraction form is just 3/1 meaning that we'll go 3 up and 1 to right (from the point you just marked) and put another point there. Now draw another big line through both of those points (WITH A RULER) and mark the ends with arrows.

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2 years ago
Use the set of ordered pairs to determine whether the relation is a one-to-one function. {(−6,21),(−23,21),(−12,9),(−24,−10),(−2
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Answer:

the relation is not one-to-one.

Step-by-step explanation:

it can't because every number is in the 4th quadrant.  

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3 years ago
Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo
noname [10]
Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;R&({{ -1}}\quad ,&{{ 6}})\quad &#10;%  (c,d)&#10;S&({{ 5}}\quad ,&{{ 5}})&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)&#10;\\\\\\&#10;\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2}  \right)}

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;&({{ -1}}\quad ,&{{ 6}})\quad &#10;%   (c,d)&#10;&({{ 5}}\quad ,&{{ 5}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}

and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\&#10;slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}&#10;&x_1&y_1\\&#10;%   (a,b)&#10;&({{ 2}}\quad ,&{{ \frac{11}{2}}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies 6&#10;\\\\\\&#10;% point-slope intercept&#10;\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)&#10;\\\\\\&#10;y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
6 0
3 years ago
Suppose that two populations have the same mean. A researcher draws a sample of size 35 from each population and calculates the
Rufina [12.5K]

Answer:

Step-by-step explanation:

Hello!

You have the difference between two sample means. Remember the sample means are variables with certain distribution, let's say, in this case, both sample means have a normal distribution. X[bar] = sample mean

X₁[bar]~N(μ₁;σ₁²/n₁)

X₂[bar]~N(μ₂;σ₂²/n₂)

Then the difference between these two variables results in a third variable that will also have a normal distribution:

X₁[bar]-X₂[bar]~N(μ₁-μ₂;σ₁²/n₁+σ₂²/n₂)

These are some of the properties of the normal distribution

Centered in μ

Symmetrical

Bell-shaped

[μ - σ; μ + σ]= 68% of the distribution

[μ - 2σ; μ+ 2σ]= 95% of the distribution

[μ - 3σ; μ+ 3σ]= 99.7% of the distribution

Check attachment.

Taking these properties into account, if you where to draw the results of the 100 trials, where μ₁-μ₂=0 would be its center and the standard deviation of the difference is 1.78.

68% of the information will be between (μ₁-μ₂) ± [(σ₁/√n₁)+(σ₂/√n₂)], this is 0 ± 1.78

I hope it helps!

3 0
3 years ago
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