All categories

3 years ago

Hey, here are all the problems. Im not sure how to do this and my teacher isnt giving me much time. Help would be apricated

Mathematics

1 answer:

ale4655 [162]3 years ago

5 0

Answer : (9x + 5) - (-8x - 11) is 17x + 16

explanation: 1st, distribute 2nd, eliminate redundant parentheses 3rd, add the numbers 4th, combine like terms.

(-5x - 7) + (-2x + 6) is -7x - 1

explanation : just do the same as every other one

(-x - 9) - (5x +4) is -6x - 13

explanation : 1st, distribute 2nd, eliminate redundant parentheses 3rd, subtract the numbers 4th, combine like terms.

(3x - 4) - (x -7) is 2x + 3

explanation : 1st, distribute 2nd, eliminate redundant parentheses 3rd, add the numbers 4th, combine like terms.

(7x + 4) - (5x + 5) is 2x - 1

explanation : 1st, distribute 2nd, eliminate redundant parentheses 3rd, subtract the numbers 4th, combine like terms.

(2x - 3) + (7x + 11) is 9x + 8

explanation : 1st, eliminate redundant parentheses 2nd, add the numbers 3rd, combine like terms.

(-4x - 5) + 2(3x - 4) is 2x - 13

explanation : 1st, distribute 2nd, eliminate redundant parentheses 3rd, subtract the numbers 4th, combine like terms.

2(6x - 7) - (2x + 3) is 10x - 17

explanation : 1st, distribute 2nd, distribute again 3rd, subtract the numbers 4th, combine like terms.

(5x + 4) - (5x + 12) is -8

explanation: 1st, distribute 2nd, eliminate redundant parentheses 3rd, subtract the numbers 4th, combine like terms.

ALSO REMEMBER YOU ARE SIMPLYFYING

You might be interested in

The copy machine at the library made a copy of Leslie's 3-page essay in just

Lesechka [4]

The answer is 100 if you do it step-by-step

6 0

3 years ago

Please help!! 10 points

hichkok12 [17]

Answer:

We conclude that:

$\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}$

Hence, option B is correct.

Step-by-step explanation:

Given the expression

$\sqrt[3]{200k^{15}}$

Apply radical rule:

$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

so the expression becomes

$\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}$

first solving

$\sqrt[3]{k^{15}}$

Apply radical rule: $\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0$

$\sqrt[3]{k^{15}}=k^{\frac{15}{3}}=k^5$

then solving

$\sqrt[3]{200}$

prime factorization: 200: 2³ · 5²

$=\sqrt[3]{2^3\cdot \:5^2}$

Apply radical rule:

$\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\sqrt[3]{2^3}\sqrt[3]{5^2}$

Apply radical rule:

$\sqrt[n]{a^n}=a,\:\quad \:a\ge 0$

$=2\sqrt[3]{5^2}$

Thus, the main expression becomes

$\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}$

$=2k^5\sqrt[3]{25}$

Therefore, we conclude that:

$\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}$

Hence, option B is correct.

3 0

3 years ago

Is the square root of 144 rational or irrational

Reptile [31]

The answer is rational

4 0

3 years ago

I need to find the zeros of the function:<br><br>f(x)=2x^2-5x-3

GaryK [48]

$2x^2-5x-3=0\\
2x^2-2x-3x-3=0\\
2x(x+1)-3(x+1)=0\\
(2x-3)(x+1)=0\\
x=\frac{3}{2} \vee x=-1$

4 0

4 years ago

Determine whether each of these sets is the power set of aset,wherea and b are distinct elements. a) ∅ c) {∅,{a},{∅,a}} b) {∅,{a

bija089 [108]

Answer:

{∅, {a}, {b}, {a,b}}

Step-by-step explanation:

The value of power of a set is generalized by using the formula,

Power of a set (P) = 2^n where n is the number of element in the set.

Given two distinct elements a and b say;

A = {a,b}

n(A) = 2 i.e the number of elements in the set is 2. Therefore the power of the set will be 2^n which gives 2^2 = 4.

P(A) = 4 means there are 4 subsets of the given set. Subsets are sets of elements that can be found in the set. The subsets of element A will be;

{∅, {a}, {b}, {a,b}} which gives 4 elements in total.

Note that empty set ∅ is always part of the subset of any given set

3 0

4 years ago