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laila [671]
3 years ago
11

At the store, Alyssa spent $3.25 on stickers and $6.50 on candy. She paid with a $10 bill. How much change did she get back?

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
7 0
A quarter or $0.25.
emmasim [6.3K]3 years ago
7 0

Answer:

$0.25

Step-by-step explanation:

First, find how much the stickers and candy cost in total:

3.25 + 6.50

= 9.75

Subtract this from 10 to find how much change she got back:

10 - 9.75

= 0.25

So, she got back $0.25 in change

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If D is the midpoint of EG, find the length of EG
Brrunno [24]

ED = x - 5 <em>given</em>

DG = 4x - 38 <em>given</em>

ED = DG <em>definition of midpoint</em>

x - 5 = 4x - 38 <em>substitution</em>

-5 = 3x - 38 <em>subtraction property of equality (subtracted x from both sides)</em>

33 = 3x <em>addition property of equality (added 38 to both sides)</em>

11 = x <em>division property of equality (divided 3 from both sides)</em>

ED = x - 5 → ED = 11 - 5 → ED = 6 <em>substitution</em>

since ED = DG, then DG = 6 <em>transitive property</em>

ED + DG = EG <em>segment addition property</em>

6 + 6 = EG <em>substitution</em>

12 = EG <em>simplified like terms</em>

Answer: 12

7 0
3 years ago
in a candy store a $5,00 jar of candy is labled 50% off .what is the idscoutn?what si the sales price of the jar of candy?
lisov135 [29]
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5 0
4 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
Please help not sure on this
ahrayia [7]
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4 0
3 years ago
Determine if line jk and line lm are parallel perpendicular or neither​
frez [133]

Answer:

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