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Oksanka [162]
2 years ago
10

Please answer this question! 30 points and brainliest!

Mathematics
2 answers:
tangare [24]2 years ago
4 0

Answer:

3.5 cm

Step-by-step explanation:

2.5 is half of 5 so you divide 7 by 2 and you get 3.5

Vinvika [58]2 years ago
3 0

Answer:

the length would be 3.5 cm

Step-by-step explanation:

I looked at the width of the reduced space. I looked at the regular photo and divided by 2 in order to get the length.

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Find the LCM and HCF of 15, 42 and 123
skelet666 [1.2K]

Answer:

LCM - 1

HCF - 3

Step-by-step explanation:

LCM - Lowest Common Factors

HCF - Highest Common Factors

____________________________________________________________

Factors of 15 - 1, 3, 5, 15

Factors of 42 - 1, 2, 3, 6, 7, 14, 21, 42

Factors of 123 - 1, 3, 41, 123

So now we know the factors of 15, 42, 123 what is lowest and highest factor we can find in every number

____________________________________________________________

I have bolded the lowest and highest common factor.

Factors of 15 - 1, 3, 5, 15

Factors of 42 - 1, 2, 3, 6, 7, 14, 21, 42

Factors of 123 - 1, 3, 41, 123

_______________________________________________

LCM - 1

HCF - 3

7 0
2 years ago
Find the product of 51 and -2.5
Elodia [21]

Answer: -127.5

Hope this helps!

3 0
3 years ago
Read 2 more answers
Regular pentagon RSTUV has sides that are half the length of regular pentagon JKLMN.
Andrew [12]

Answer:

c

Step-by-step explanation:

7 0
2 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
Turn upside down but somebody HELP!
bogdanovich [222]
Part of the answer is covered up.

Get  a better picture please

8 0
3 years ago
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