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zvonat [6]
3 years ago
11

How would you change the equation in the explore activity to model -3x+4=5

Mathematics
1 answer:
MissTica3 years ago
8 0
Are you asking how to solve this equation?
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There below! Can somebody help me please?
Montano1993 [528]

Answer: C. 24 units

Step-by-step explanation:

The circumference of a circle is 2πr.  Let's say x = 1/3r.  Thus, the area of circle A relative to circle B is 1:3.  Thus, circle B is 3 times larger than circle A.  So, simply divide 64π by 4 to get 16π.  Then multiply 16π*3=48π.  For the circumference of a circle to be 48, the radius would have to be 24.

Hope it helps <3

4 0
3 years ago
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What is the value of x pls help
photoshop1234 [79]

Answer:

118

Step-by-step explanation:

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Angie and Kenny play online video games. Angie buys 1 software package and 3 months of game play. Kenny buys 2 software packages
irga5000 [103]

Answer:

$10

Step-by-step explanation:

cost of one month of game play = x

1(35) + 3x + 2(35) + 5x = 185

35 + 70 + 8x = 185

105 + 8x = 185

-105           -105

8x = 80

/8.     /8

x =10

7 0
1 year ago
3. Jim bought a new pair of glasses. The lenses had six sides. The lenses were in the shape of a A. hexagon. B. heptagon. C. qua
Vladimir [108]
Best Answer:<span>  </span><span>6 sides is a hexagon. 

the spoke of a wheel is like a radius. Circumference is equal to 2πr so r = 42/2π ≈ 6.69 inches, ans. D. 

that's it! ;)</span>
5 0
3 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
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