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krek1111 [17]
3 years ago
13

Kelsey has saved $35 towards the cost of A $390 video game in the weeks to come she plans to save in additional $10 per week use

unit analysis to represent them out just saved after W weeks and identify the units for the expression
Mathematics
2 answers:
LenKa [72]3 years ago
3 0

390-35=355

355-10w

355/10=35.5

355-10(35.5)

35 and a 1/2 weeks

36 weeks to have the full amount needed

loris [4]3 years ago
3 0

Answer:

36th weeks

Step-by-step explanation:

Kelsey has saved $35 to buy a video game worth $390.

She plans to save additional $10 per week.

Let w represents number of week.

Let s represents total saving in w weeks.

Addition saving per week = $10

In w weeks saving  = 10w

Previous saving amount = $35

Total saving in w week, s = 35 + 10w

Expression which represents total saving in w weeks

s = 35 + 10w

Now find number of weeks to save $390

So, put s=390 and solve for w

35 + 10w = 390

        10w = 390 - 35

           w = 355÷10

          w = 35.5

Hence, Kelsey will save $390 in 36th weeks.

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Which ordered pair would form a proportional relationship with the point graphed below?
erastovalidia [21]

Answer:

(10, -20)

Step-by-step explanation:

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3 years ago
There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many
svp [43]

Answer:

Option A - 10

Step-by-step explanation:

Given : There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB.

To find : How many color patterns are possible?

Solution :

Total number of chips = 5

So, 5 chips can be arranged in 5! ways.

There are 3 red chips and 2 blue chips.

So, choosing 3 red chips in 3! ways

and choosing 2 blue chips in 2! ways.

As changing the places of similar chip will not create new pattern.

The total pattern is given by,

T=\frac{5!}{3!\times 2!}

T=\frac{5\times 4\times 3!}{3!\times 2}

T=10

Therefore, color patterns are possible are 10.

Option A is correct.

4 0
3 years ago
Will give 20 points
beks73 [17]

Answer:

the firtst part is 2  2

c11 =  -1

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c21 =  3

c22 = -3

the next part is

d11 =   0

d12 =  6

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Step-by-step explanation:

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7 0
4 years ago
Read 2 more answers
Please help and show work
oee [108]

Answer:

Coordinates of A': will NOT be the same

Coordinates of C': will NOT be the same

Perimeter of ABC: will be the same

Area of A'B'C': will be the same

Measure of ∠B: will be the same

Slope of A'C': will be the same

Step-by-step explanation:

We have a triangle ABC that will be translated (moved) 2 units down, 3 units to the right.

Since the triangle is moved, then the coordinates of every summit (A, B and C) will be affected.  So,

Coordinates of A': will NOT be the same

Coordinates of C': will NOT be the same

However, since the triangle is only moved, not transformed in any way, not scaled up/down for example the following will remain the same:

Perimeter of ABC: will be the same

Area of A'B'C': will be the same

Measure of ∠B: will be the same

Since the triangle is only translated, not rotated or reflected the following will remain the same:

Slope of A'C': will be the same

3 0
4 years ago
Solve the given inequality :
tigry1 [53]

Answer:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

Step-by-step explanation:

Given <u>rational inequality</u>:

\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0

\textsf{Factor }(x^2-5x+6):

\implies x^2-2x-3x+6

\implies x(x-2)-3(x-2)

\implies (x-3)(x-2)

Therefore:

\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0

Find the roots by solving f(x) = 0  (set the numerator to zero):

\implies (x-1)^2(x-2)^3=0

\implies (x-1)^2=0\implies x=1

\implies (x-2)^3=0 \implies x=2

Find the restrictions by solving f(x) = <em>undefined  </em>(set the denominator to zero):

\implies (x-3)^2(x-2)^2(x+5)=0

\implies (x-3)^2=0 \implies x=3

\implies (x-2)^2=0 \implies x=2

\implies (x+5)=0 \implies x=-5

Create a sign chart, using closed dots for the <u>roots</u> and open dots for the <u>restrictions</u> (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values:  -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+

f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

Record the results on the sign chart for each region (see attached).

As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).

Therefore, the solution set is:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

As interval notation:

(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)

4 0
2 years ago
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