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Savatey [412]
2 years ago
6

WILL MARK BRAINLIEST

Mathematics
1 answer:
Rina8888 [55]2 years ago
8 0

Answer:

the answer is yes

Step-by-step explanation:

trust me

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How would you determine mike's income?
LUCKY_DIMON [66]
Good I am pretty sure I am a smart person so….
8 0
2 years ago
I got 10 min to submit pls I need it asap
babymother [125]

Answer:

We just need the equations to equal the same, which is y.

5+1.10x=y

3+1.50x=y

5+1.10x= 3+1.50x

Let's get x by itself by subtract 3 from both sides.

2+1.10x= 1.50x

Now subtract 1.10x from both sides.

2= 0.4x

Now divide each side by 0.4

5=X

So, at 5 games they're the same cost. Let's find the cost.

5+1.1(5)= 5+5.5=10.50

3+1.5(5)=3+7.5= 10.50

So, at 5 games, they are $10.50.

Hope this helps! :)

5 0
2 years ago
I need help plz!!?!???!?
Ksivusya [100]

Answer:

Solution → (3, -1)

Step-by-step explanation:

We ave to identify the graph representing the system of lines first.

y = \frac{2}{3}x-3

Y-intercept of the line = -3

y = -2x + 5

y-intercept of the line = 5

From the given graphs 3rd (extreme right) graph is representing the system of equations.

Solution of the system of equations will be the point of intersection of these lines.

Solution of the system → (3, -1)

8 0
2 years ago
What is the equation of this line in standard form? (-4, -1) & (1/2, 3)
BabaBlast [244]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-4)}}}\implies \cfrac{3+1}{\frac{1}{2}+4}\implies \cfrac{4}{~~\frac{9}{2}~~}\implies \cfrac{4}{1}\cdot \cfrac{2}{9}\implies \cfrac{8}{9}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{8}{9}}[x-\stackrel{x_1}{(-4)}]\implies y+1=\cfrac{8}{9}(x+4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{9}}{9(y+1)=9\left( \cfrac{8}{9}(x+4) \right)}\implies 9y+9=8(x+4)\implies 9y+9=8x+32 \\\\\\ 9y=8x+23\implies -8x+9y=23\implies 8x-9y=-23

4 0
2 years ago
the admission fee for a charity event is 7 dollars for children and 10 dollars for adults the event was attended by 700 people a
ollegr [7]
Given:
$7/child
$10/adult
Total people = 700
Total money = $6,400

First, make two equations.
Let a = # of adults & Let c = # of children.
Let p = total people

1.      a+c = 700
2.  10a+7c = 6,400

Then, rearrange the equation to solve for a variable.
c = 700-a

Substitute (700-a) for c, or the # of children in the second equation.
10a+(700-a) = 6400
9a+700 = 6400
9a+700-700 = 6400-700
9a = 5700
9a/9 = 5700/9 
a = 633\frac{1}{3}  = # of adults attended

700-633\frac{1}{3} = c = 66 \frac{2}{3} = # of children attended


4 0
3 years ago
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