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rosijanka [135]
3 years ago
8

What steps do you use to solve a linear inequality

Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0
First you need to isolate the term X variable on one side of the equation by either subtracting to combine like terms
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What is the area of this triangle? Enter your answer in the box. units² ​
Vesnalui [34]

Answer:

10u²

Step-by-step explanation:

The formula for area of a triangle is b*h/2. When you plug in the numbers, you get 10. Hope this helps!

6 0
1 year ago
Please help asap.<br> I will give brainliest
alexandr1967 [171]

Answer:

#1 20

#2 17

#3 14

#4 14

#5 12

#6 14

#7 10

#8. 20

#9. 20

Step-by-step explanation:

1 12+8=20

2 10+7=17

3 8+6=14

4 8+6=14

5 10+2=12

6. 8+6=14

7 4+6=10

8. 12+8=20

9 14+6=20

Now honey I can't do all your work for you im.not trying to be mean but work some.out heres a tip use a calculator to help you when you get stuck ask me your question and then I will hold you ok

5 0
1 year ago
The main mathematical step used to solve percent proportion is...
Scrat [10]

To solve percent problems, you can use the equation, Percent · Base = Amount, and solve for the unknown numbers. Or, you can set up the proportion, Percent =, where the percent is a ratio of a number to 100. You can then use cross multiplication to solve the proportion.

Hope this Helps you... =)=)

6 0
3 years ago
Read 2 more answers
The fourth term of an AP is 13 and the second term is 3 find the common difference<br><br>​
kogti [31]

Answer:

5

Step-by-step explanation:

<h3>Given</h3>

<u>In and AP</u>

  • a4 = 13
  • a2 = 3
  • d = ?
<h3>Solution</h3>

<u>Formula for the terms of AP:</u>

  • a4 = a + 3d
  • a2 = a + d

<u>The difference of terms:</u>

  • a + 3d - (a - d) = 13 - 3
  • 2d = 10
  • d = 5
3 0
3 years ago
For the level 3 course, exam hours cost twice as much as workshop hours, workshop hours cost twice as much as lecture hours. How
natulia [17]
<h2>Answer</h2>

Cost of lectures = $7.33 per hour

<h2>Explanation</h2>

Let e the cost of the exam hours

Let w be the cost of the workshop hours

Let l be the cost of the lecture hours.

We know from our problem that exam hours cost twice as much as workshop, so:

e=2w equation (1)

We also know that workshop hours cost twice as much as lecture hours, so:

w=2l equation (2)

Finally, we also know that 3hr exams 24hr workshops  and 12hr lectures cost $528, so:

3e+24w+12l=528 equation (1)

Now, lets find the value of l:

Step 1.  Solve for l in equation (3)

3e+24w+12l=528

12l=528-3e-24w equation (4)

Step 2. Replace equation (1) in equation (4) and simplify

12l=528-3e-24w

12l=528-3(2w)-24w

12l=528-6w-24w

12l=528-30w equation (5)

Step 3. Replace equation (2) in equation (5) and solve for l

12l=528-30w

12l=528-30(2l)

12l=528-60l

72l=528

l=\frac{528}{72}

l=\frac{22}{3}

l=7.33

Cost of lectures  = $7.33 per hour



3 0
3 years ago
Read 2 more answers
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