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Umnica [9.8K]
3 years ago
12

Write an inequality and solve each problem.

Mathematics
1 answer:
astra-53 [7]3 years ago
5 0
For all of these, let n = the unknown number. 

25) "Five less than a number" becomes 5 - n 
"At least" tells you your number is greater than or equal to - 2.
Put it all together.

5 - n ≥ - 2
- n ≥ - 2 - 5
- n ≥ - 7
Divide by - 1 to isolate the variable. This flips the operation.

n ≤ 7

26) "The difference between a number and 6" tells you to do n - 6
"no more than" means this is a less than or equal to sign. 

n - 6 ≤ 5
n ≤ 5 + 6

n ≤ 11

27) "The sum of a number and 7" becomes n + 7
"more than" is a greater than sign

n + 7 > 1
n > 1 - 7

n > - 6

28) "The difference between a number and 10" becomes n - 10
Obvious one, "greater than" is... just that.

n - 10 > 9
n > 9 + 10

n > 19

29) "Four less than a number" is 4 - n
Another obvious one in "less than" 

4 - n < 11
- n < 11 - 4
- n < 7
Divide by - 1 on both sides to isolate the variable

n > - 7
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The graph below shows the solution to which system of inequalities
Oksi-84 [34.3K]
The answer is
the option A
<span>The graph shows the solution of the system of inequalities
y> -3
y<=-x

using a graph tool
see the attached figure</span>

5 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP -- WITH THE WORKK<br><br>50 points
MAXImum [283]

Part-A:-

<h3>(f×g)=</h3>

\\ \sf\longmapsto x^2+3x-70(4x-12)

\\ \sf\longmapsto 4x^2+12x-280-12x^2+36x+840

\\ \sf\longmapsto -8x^2+48x+560

Now

\\ \sf\longmapsto (f.g)(3)

\\ \sf\longmapsto -8(3)^2+48(3)+560

\\ \sf\longmapsto -72+144+560

\\ \sf\longmapsto 72+560

\\ \sf\longmapsto 632

Part-B:-

\\ \sf\longmapsto f(3)

\\ \sf\longmapsto (3)^2+3(3)-70

\\ \sf\longmapsto 9+9-70

\\ \sf\longmapsto 18-70

\\ \sf\longmapsto -52

Now

\\ \sf\longmapsto g(f(3))

\\ \sf\longmapsto g(-52)

\\ \sf\longmapsto 4(-52-3)

\\ \sf\longmapsto 4(-55)

\\ \sf\longmapsto -220

Part:-C

\\ \sf\longmapsto g(-2)

\\ \sf\longmapsto 4(-2-3)

\\ \sf\longmapsto 4(-5)

\\ \sf\longmapsto -20

Now

\\ \sf\longmapsto f(g(-2))

\\ \sf\longmapsto f(-20)

\\ \sf\longmapsto (-20)^2+3(-20)-70

\\ \sf\longmapsto 400-60-70

\\ \sf\longmapsto 400-130

\\ \sf\longmapsto 270

Part-D:-

  • g×f=f×g

\\ \sf\longmapsto (g.f)(-2)

\\ \sf\longmapsto -8(-2)^2+48(-2)+560

\\ \sf\longmapsto 32-96+560

\\ \sf\longmapsto -64+560

\\ \sf\longmapsto 496

6 0
2 years ago
Pine Bluff Middle School is having its annual Spring Fling dance, which will cost
Anon25 [30]
I think your asking  how much the school will have left over. If so:

125 ( not including the tickets)
It depends on how many kids are coming to the dance in order to fully solve this problem.

You didn't word this properly so that what I got with what I had:)

If you have anymore questions, ask me them on my profile so I'll be sure to get them:)

I hope this helps:)
3 0
3 years ago
Mrs. Santos is buying binders for the students in her class. She determined that 15 binders would cost $22.50.
atroni [7]

Answer:

It will cost $42 for 28 binders

Step-by-step explanation:

We know that 15 binders = 22.50 dollars . We need to find the cost of 18, 30, 28, and 48 binders to figure out if they match up with the statements. One way to do this is to find the cost for one binder, and then multiply that by the number of binders to find the cost for a certain number of binders. For example, if one binder costs 2 dollars, five binders would cost 2 * 5 = 10 dollars.

To figure out how much one binder costs, we can use the division property of equality to divide both sides of our equation (15 binders = 22.50 dollars) by 15. We divide by 15 because anything divided by itself is equal to 1, and that would make it 1 binder = something dollars.

Applying this, we get

(15 binders)/15 = (22.5 dollars)/15

1 binder = 1.5 dollars

For 18 binders, this would cost 1.5 * 18 = 27 dollars. We know this because we add 1.5 dollars for each binder.

For 30 binders, this would cost 30*15=45 dollars

For 28 binders, this would cost 1.5 * 28 = 42 dollars

For 48 binders, this would cost 1.5*48 = 72 dollars

The only one that matches up is 28 binders for 42 dollars

8 0
3 years ago
Consider the proof. Given: In △ABC, BD ⊥ AC Prove: the formula for the law of cosines, a2 = b2 + c2 – 2bccos(A) Statement Reason
loris [4]

Solution: The missing reason in Step 8 is substitution of x=c\cos (A).

Explanation:

The given steps are used to prove the formula for law of cosines.

From step 5 it is noticed that our equation is

a^2=b^2-2bx+c^2      ..... (1)

From step 7 it is noticed that the value of x is c\cos (A).

So by substituting c\cos (A) for x in equation (1) we get the equation of step 8, i.e.,

a^2=b^2-2bc\cos (A)+c^2

Hence, the missing reason in Step 8 is substitution of x=c\cos (A).



5 0
3 years ago
Read 2 more answers
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