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Ad libitum [116K]

3 years ago

14

11+2 to the 3rd power ×5 using PEMDAS?

2 answers:

uysha [10]3 years ago

8 0

11+2^3x5

You always start with parenthesis but there aren't any so you go to exponents. 2 to the third power is 2 x 2 x 2 which is 8. so now you will have 11+8x5. Next is either multiply or divide left to right. There's only multiplication so you have to multiple 8 x 5 which is 40. now all you're left with is 11 + 40. You have to add or subtract from left to right. Since, there's only adding, 11 + 40 = 51

allochka39001 [22]3 years ago

6 0

58 - its simple- Hope I helped :)

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VARVARA [1.3K]

When solving an equation with an absolute value term, you make two separate equations ans solve for x:

Equation 1: |4x-3|-5 = 4

1st add 5 to both sides:

|4x-3| = 9

Remove the absolute value term and make two equations:

4x-3 = 9 and 4x - 3 = -9

Solving for x you get X = 3 and x = -1.5

When you replace x with those values in the original equation the statement is true so those are two solutions.

Do the same thing for equation 2:

|2x+3| +8 = 3

Subtract 8 from both sides:

|2x+3| = -5

Remove the absolute value term and make two equations:

2x +3 = -5

2x+3 = 5

Solving for x you get -1 and 4, but when you replace x in the original equation with those values, the statement is false, so there are no solutions.

The answer is:

C. The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution.

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3 years ago

For the function y = 7 - 3x, what is the ordered pair of x=5?

Alexeev081 [22]

Answer:

9

Step-by-step explanation:

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2 years ago

Read 2 more answers

Please help me 33/2 + 3y/5 = 7y/10 + 15

8090 [49]

We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

${:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}$

Take LCM on both sides :

${:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}$

<em>Multiplying</em> both sides by <em>10</em> ;

${:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}$

${:\implies \quad \sf 165+6y=7y+150}$

Can be <em>further written</em> as ;

${:\implies \quad \sf 7y+150=165+6y}$

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

${:\implies \quad \sf 7y-6y=165-150}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}$

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avanturin [10]

Answer:

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Step-by-step explanation:

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