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3 years ago

Simplify the following expression. 2p – 10 – 4p – 7

Mathematics

2 answers:

aev [14]3 years ago

8 0

Answer:

2p – 10 – 4p –7

–2p –17

I hope I helped you^_^

Roman55 [17]3 years ago

3 0

Answer:

the answer is-2p-3 

hope is helpful

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konstantin123 [22]

Answer:

a= 7.5%12.58= 0.9435 round 0.94+12.58= 13.52

b= 31.50 divided by 19.00= 1.65789474 round 1.66

c= (14.25 x 18)+ (5.5%250)

l l

v v

256.50 + 13.75 = 270.25

Step-by-step explanation:

5 0

3 years ago

after a school election,Jamie had 3/8 of the votes Richard the only other candidate had 75 votes how many people voted for jaime

dalvyx [7]

Jamie and Richard are the only contestants of the election.

Jamie got 3/8 of the total votes.

Richard got = 1 - 3/8 = (8-3)/8 = 5/8 of total votes

Number of votes Richard got = 75

So 5/8 of total votes = 75

Total votes = 75 ÷ 5/8 = 75 × 8/5 = 120

Jamie got = 120 - 75 = 45 Votes

8 0

4 years ago

A rhombus can belong to multiple categories. which catergories does this rhombus not belong to? Select the two correct answers.

amid [387]

Answer:

Im pretty sure its c and d

6 0

3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago

Four times the larger of two numbers is equal to seven times the smaller. The sum of the number is 22. Fine the number

GenaCL600 [577]

4y=7x
The sum of the numbers is 22
so, x+y=22
x=(22-y)
by substitute (22-y) for x in the first equation.

6 0

3 years ago