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3 years ago

Solve the equation 1 + 4 (2x + 1) = 5

Mathematics

2 answers:

Marizza181 [45]3 years ago

7 0

Answer:

2x+1= 1+4•(3x)= 12x+1 = 13x

Natali [406]3 years ago

6 0

Answer:

x=0

Step-by-step explanation:

1+4(2(0)+1)=5

1+4=5

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Answer the following

Amanda [17]

The set A satisfying the given inequality is A = (- $\infty$ , -10].

<h3>What are some properties of an inequality relation? </h3>

Following are some facts which are true for an inequality relation:

Equal numbers can be added or subtracted from both sides of an inequality without affecting the inequality sign.
The Inequality sign is unchanged if both sides are multiplied or divided by a positive number, but when multiplied or divided by a negative number the inequality sign is reversed.

$\frac{5x-2}{8} - \frac{3x-5}{10} &\ge& x+y\\\\\Rightarrow\;\; \frac{13}{40}x + \frac{1}{4}&\ge& x+y\\\\\Rightarrow\;\;\;\;\;\; -\frac{27}{40}x &\ge & y - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\; -x &\ge & \frac{40}{27}\left( y-\frac{1}{4} \right).\hspace{1cm}(1)$

Since y ∈ B, -2 ≤ y ≤ 7. So,

$\;\;\;\;\;\;\;\,-2 - \frac{1}{4}\; \le\; y - \frac{1}{4} \;\le\; 7 - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\; -\frac{9}{4}\; \le\; y - \frac{1}{4} \;\le\; \frac{27}{4}\\\\\Rightarrow\;\;\; -\frac{9}{4}\cdot \frac{40}{27} \;\le\; \frac{40}{27} \left( y-\frac{1}{4} \right) \;\le \;\frac{27}{4}\cdot \frac{40}{27}\\\\\Rightarrow\;\;\;\;\;\;\;\; -\frac{10}{3} \;\le\; \frac{40}{27}\left( y - \frac{1}{4} \right)\; \le\; 10.$

The set {-x | inequality (1) holds ∀ y ∈ B} is [10, $\infty$ ) i.e.

10 ≤ -x ≤ $\infty$ .

Multiplying -1 throughout gives

-10 ≥ x ≥ - $\infty$ .

x, thus, lies in the range A = (- $\mathbf{\infty}$ , -10}.

Learn more about the inequality here.

brainly.com/question/17801003

Disclaimer: The question was incomplete. Please find the full content below.

<h3>Question </h3>

Find the set A such that for x ∈ A

$\frac{5x - 2}{8} - \frac{3x - 5}{10} \ge x + y$

∀y ∈ B = {y ∈ R | -2 ≤ y ≤ 7}.

#SPJ4

4 0

2 years ago

Is the sum of two monomials always a monomial? Is their product always a monomial?

Ivan

Sum of two monomials is not necessarily always a monomial.

For example:

Suppose we have two monomials as 2x and 5x.

Adding 2x+5x , we get 7x.

So if two monomials are both like terms then their sum will be a monomial.

Suppose we have two monomials as 3y and 4x

Now these are both monomials but unlike, so we cannot add them together and sum would be 3y + 4x , which is a binomial.

So if we have like terms then the sum is monomial but if we have unlike terms sum is binomial.

Product of monomials:

suppose we have 2x and 5y,

Product : 2x*5y = 10xy ( which is a monomial)

So yes product of two monomials is always a monomial.

4 0

3 years ago

Read 2 more answers

Please help me with this question!! Thank you so much, it should be quick. I need to know what goes in the blank space above.

Ipatiy [6.2K]

If two objects have the same shape, they are called "similar." When two figures are similar, the ratios of the lengths of their corresponding sides are equal so the word would be ratio

3 0

4 years ago

Read 2 more answers

Lines m and n are parallel. They are cut by transversal t. What other angle is equal to 65 degrees?

Marizza181 [45]

Angle 3, they are opposite interior angles

5 0

4 years ago

Read 2 more answers

Seven lamps labeled A through G are arranged in a row. Each lamp has its own switch. Now lamps A, C, E, and G are on and other l

Yanka [14]

Answer:

O=on

X=off

OXXOXOX after 2011 flips.

Step-by-step explanation:

There are many ways to get the answer.

A. Actually flip them 2011 times.

One way is to put 7 cards on the table, and name them A to G, with A,E,C,G facing up, and the other facing down.

Now flip the cards in order A to G 2011 times and find the answer.

Since you have four days to find the answer, you only have to flip about 500 times a day.

B. Simplify procedure A above.

Do the same as above, actually flip the cards, but only 14 times.

You will find that the cards are restored to the original position after 14 flips.

If we divide 2011 by 14, we get a quotient of 143 with a remainder of 9.

We know we don't have to make 143 sets of 14 flips. We only need to make 9!

Then again, we know that after 7 flips, the lights will be all opposite.

If O=on, and X=off then the situations are

OXOXOXO initial

XOXOXOX after 7 flips

OXXOXOX after 9 flips, which is also after 2011 flips, and that is the final answer.

7 0

4 years ago