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uysha [10]
4 years ago
13

Help asap my last question GIVING BRAINLIST!!!!

Mathematics
1 answer:
Alex4 years ago
6 0

Answer:

1) x < 0

2) x > 0

3) x < 10

4) x < -\frac{31}{55}

5) x \geq 0

6) x \geq -1

Step-by-step explanation:

1) -3 > 2x -3

2x- 3 < -3

2x - 3 + 3 < -3 +3

2x < 0

\frac{2x}{2} < \frac{0}{2}

x < 0

2) 5 > -x + 5

-x + 5 < 5

- x + 5 -5 < 5 - 5

-x < 0

\frac{-x}{-1} < \frac{0}{-1}

x > 0

3) \frac{x}{5} -2 < 0

5 \times (\frac{x}{5} - 2 ) \times 5 < 0

x - 10 < 0

x -10 + 10 < 0 +10

x < 10

4) -5x + 11 > 31

-5x \cdot 11 < 31

-55x < 31

\frac{-55x}{-55} > \frac{31}{-55}

x < -\frac{31}{55}

5) \frac{1}{2}x - 4 \geq -4

\frac{1}{2}x -4 +4 \geq -4 +4

\frac{1}{2}x \geq 0

\frac{\frac{1}{2}x }{\frac{1}{2} } \geq \frac{0}{\frac{1}{2} }

x \geq 0

6) -4x + 2 \leq 6

-4x +2 - 2 \leq 6 -2

-4x < 4

\frac{-4x}{-4} \leq  \frac{4}{-4}

x \geq -1

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Sara has 3000 to invest for 2 years she invest the money in a bank that pays simple interest at the rate of 7.5 per year calcula
gavmur [86]

Answer:

450

Step-by-step explanation:

A = P(1 + rt)

A = final amount

P = initial principal balance

r = annual interest rate

t = time(in years)

Given,

P = 3000

t = 2 years

r = 7.5% = 7.5/100 = 0.075

A = 3000(1 + 0.075*2) = 3450

Interest = 3450 - 3000 = 450

8 0
3 years ago
Grant went on a eight week diet if you lost at least 1.5 points each week which inequality gives his total weight loss
andrezito [222]

Answer:

12 in total    8<1.5x

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
How do you solve these two equations? Please Help!!<br>y = 2x + 160 <br>y = - 3x + 208
Archy [21]
Their are 2 main methods of solving 2 equations with 2 variables:
1) SUBSTITUTION
2) ELIMINATION
I'll show you both:

1) y = 2x + 160
y = - 3x + 208
plug what y is equal to in 1st equation into the y of the 2nd:
2x + 160 = -3x + 208
2x + 3x + 160 = -3x + 3x + 208
5x + 160 -160 = 208 - 160
5x = 48
5x/5 = 48/5
x = 48/5 = 9 3/5
now plug that x into either one of the original equations: y = -3x + 208 = -3(48/5) + 208
y = -144/5 + 208 = 1040/5 - 144/5 = 896/5
y = 179 1/5

2) add the whole equations together so as to cancel out one of the variables:
I'm choosing to eliminate the y, so we need y-y to remove the y's, I'll multiply the whole 2nd equation by -1 to make that -y
-1 (y = - 3x + 208)
-y = 3x - 208
So set it up like:
y = 2x + 160
-[y = 3x - 208]
---------------------
0 = 5x - 48
0+48 = 5x -48 +48
5x = 48
5x/5 = 48/5
x = 48/5
plug that x value into one of the original equations: y = 2x + 160 = 2(48/5) + 160
y = 96/5 + 160 = 96/5 + 800/5 = 896/5
y = 179 1/5

SO WE GOT THE SAME VALUES BOTH WAYS.
WHEN YOU SEE FUTURE PROBLEMS LIKE THESE, SOME WILL BE QUICKER BY ELIMINATION AND OTHERS BY SUBSTITUTION

6 0
4 years ago
A farmer has 300 ft of fencing with which to enclose a rectangular pen next to a barn. The barn itself will be used as one of th
blondinia [14]
This question can be answered in a number of ways depending on your level of mathematical sophistication (using calculus, graphing a parabola, and guess/check). I'll talk through all three in the order above.

They all start the same way: Let's call the length of the rectangle b and the width a. The perimeter of the rectangle (the distance all the way around -- that is, the length of the "fence") is 2a + 2b. Since the barn is used as one of the sides (let's say b) we can subtract b ... we don't need fencing for this side. That makes the perimeter 2a + b. Since we have 300 feet of fencing we set these equal: 2a + b = 300.

We can solve for b above by subtracting 2a from both sides of the equation so we get: b = 300 - 2a.

We are asked to maximize the area. The area of a rectangle is length times width. In this case, (a)(b) and since we just found an expression for b, we can substitute this to obtain: Area = (a)(300-2a)=300a-2 a^{2}. This is a parabola when graphed and we are asked to find where it's highest point occurs because this is where the area is greatest.

METHOD 1: Using Calculus
We are looking for the highest point of the equation Area = [tex](a)(300-2a)=300a-2 a^{2}. Let's call the function R(a) since it is in terms of a. To find the maximum we take the first derivative, set it equal to zero and solve for a. As this is a parabola (with the leading coefficient being negative) it is concave down and the point we find will be the only maximum.

First we find the derivative of the function.
R(a)=(a)(300-2a)=300a-2 a^{2} [tex]\\Next set it equal to 0 and solve for a. R^{'}(a)=300-4a&#10;0=300-4a&#10;-300=-4a&#10;75 = a [/tex]
This means that the maximum value (the largest area) occurs when a = 75.
At a = 75 the area is given by
What is the largest area? Let's us substitute 75 for a in the equation of the area. This gives: Area=a(300-2a)=(75)(300-(2)(75))=(75)(150)=11250 square feet.

METHOD 2: Graphing
We know that the area is given by. Area = (a)(300-2a)=300a-2 a^{2}. To make this more familiar let's use y for the area and x for the width (a). That means we have the equation: (a)(300-2a)=300a-2 a^{2}.

This is a quadratic equation (the highest exponent of x is 2) and so the equation of a parabola (like the letter U or an upside down U). Since the leading coefficient is negative (-2) we know it opens downward (called concave down -- think of a frown). We are looking for the x-coordinate of the highest point called the vertex. We can find this by remembering that the x-coordinate of the vertex is given by x= \frac{-b}{2a} where a is the leading coefficient the coefficient of the term with the x squared (here = -2) and a is the coefficient of the linear term (the one with x...here 300). That makes the x-coordinate of the vertex x= \frac{-300}{2(-2)}= \frac{-300}{-4}=75.

The y-coordinate of the vertex is found by substituting 75 for x in the equation for the area. y=300x-2 x^{2} &#10;y=300(75)-2(75)^{2} &#10;&#10;
y=22500- 2(5625)&#10;y=22500- 11250&#10;y=11250

Since we let x be the width of the rectangle. We know that the width is 75. The area of the rectangle we called y. So the maximum area is 11250 square feet which occurs when the width of the rectangle is 75.

You could have gotten to this point by generating a list of points for the parabola and graphing it carefully on graph paper (though it might not have been exact depending on how you scale your axis).

METHOD 3: Guess and Check
The area of the rectangle is given by Area = (a)(300-2a)=300a-2 a^{2}. Here you guess values for a, plug them into the formula to get the area and keep doing this until you are comfortable you have found the largest area. Not the best method but if it's a test and you forget everything else, it might get you close to an answer or even get you the exact answer (depending on how well you guess). It might take a while so I'd do this question last if this were a test.
 

6 0
3 years ago
6. If b&gt;6, then 3b + 3 could be<br> A. 18<br> B. 19<br> C. 20<br> D. 21<br> E. 22
Gala2k [10]

Answer:

E. 22

Step-by-step explanation:

Hello!

We know that b>6, and we are asked to find the possible values for 3b + 3.

Let's directly substitute 6 for b:

  • 3(6) + 3
  • 18 = 3
  • 21

Now, because b is greater than 6, and not equal to 6, any value greater than 21 will work.

The correct option is E. 22.

7 0
2 years ago
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