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3 years ago

What is the inequality to 2w+5>45

Mathematics

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

w > 20

Step-by-step explanation:

Perhaps you want the solution to the inequality 2w+5>45.

Subtract 5:

2w > 40

Divide by 2:

w > 20 . . . . this is the solution

_____

Comment on solving inequalities

An inequality is solved in virtually the same way that an equation is solved: undo what is done to the variable. The only difference is that when you perform an operation that reverses the ordering, you need to reverse the ordering symbol.

Multiplying or dividing by a negative number will reverse the ordering. Consider, for example, ...

2 > 1

Multiplying both sides by -1, we get ...

-2 < -1 . . . . . the > sign needs to be reversed to a < sign to keep the inequality true

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What is y=4/5x+4 in standard form

HACTEHA [7]

5y = 4x + 20
- 4x + 5y = 20

5 0

2 years ago

Rearrange the formula to make a the subject: B = 1/2h(a+b)

Mkey [24]

Answer:

Step-by-step explanation:

B=2h(a+b) #the numerator goes to the other side, the denominator then just makes it B

B/2h=a+b

2h=a+b/B

2h-a=b/B

2h-a=b(1) #factorise

6 0

3 years ago

Simplify the following expression as much as you can use exponential properties. (6^-2)(3^-3)(3*6)^4

gtnhenbr [62]

Answer:

Simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

Step-by-step explanation:

We need to simplify the expression $(6^{-2})(3^{-3})(3*6)^4$

Solving:

$(6^{-2})(3^{-3})(3*6)^4$

Applying exponent rule: $a^{-m}=\frac{1}{a^m}$

$=\frac{1}{(6^{2})}\frac{1}{(3^{3})}(18)^4\\=\frac{(18)^4}{6^{2}\:.\:3^{3}} \\$

Factors of $18=2\times 3\times 3=2\times3^2$

Factors of $6=2\times 3$

Replacing terms with factors

$=\frac{(2\times3^2)^4}{(2\times 3)^{2}\:.\:3^{3}} \\=\frac{(2)^4\times(3^2)^4}{(2)^2\times (3)^{2}\:.\:3^{3}} \\$

Using exponent rule: $(a^m)^n=a^{m\times n}$

$=\frac{(2)^4\times(3)^8}{(2)^2\times (3)^{2}\:.\:3^{3}} \\=\frac{2^4\times 3^8}{2^2\times 3^{2}\:.\:3^{3}}$

Using exponent rule: $a^m.a^n=a^{m+n}$

$=\frac{2^4\times 3^8}{2^2\times 3^{2+3}}\\=\frac{2^4\times 3^8}{2^2\times 3^{5}}$

Now using exponent rule: $\frac{a^m}{a^n}=a^{m-n}$

$=2^{4-2}\times 3^{8-5}\\=2^{2}\times 3^{3}\\=4\times 27\\=108$

So, simplifying the expression $(6^{-2})(3^{-3})(3*6)^4$ we get $\mathbf{108}$

7 0

3 years ago

"The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N obj

Liula [17]

Options

Counting rule for permutations
Counting rule for multiple-step experiments
Counting rule for combinations
Counting rule for independent events

Answer:

(C)Counting rule for combinations

Step-by-step explanation:

When selecting n objects from a set of N objects, we can determine the number of experimental outcomes using permutation or combination.

When the order of selection is important, we use permutation.

However, whenever the order of selection is not important, we use combination.

Therefore, The counting rule that is used for counting the number of experimental outcomes when n objects are selected from a set of N objects where order of selection is not important is called the counting rule for combinations.

7 0

3 years ago

Your cell phone bill in August was $61.43, which was $21.75 more than your bill in July. Your cell phone bill in July was $13.62

irina [24]

Answer:

$53.30

Step-by-step explanation:

61.43-21.75=39.68 is the cell phone bill in July.

39.68+13.62= 53.30 is the phone bill in June.

3 0

3 years ago

Read 2 more answers