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3 years ago

PLEASE HELP FAST!! |x+5|>-1, what is x (answer in x equals to)

1 answer:

7 0

$(x+5)>-1 or -(x+5)>-1$

$x>-1-5 or x$

$x>-6 or x$

This gives us the union of two sets

$(-∞,-4) U (-6,+∞)$

$=(-∞,+∞)$

x belongs to all real numbers

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Arliss has two pieces of carpet runner. One is yards long and the other is yards long. She needs 10 yards of carpet runner altog

Alekssandra [29.7K]

Given:
10 yards required
5 2/3 yards on hand.

We need to subtract the yards on hand from the total yards required.

First, we need to convert the mixed fraction into an improper fraction.

5 2/3 = ((5*3)+2)/3 = (15+2)/3 = 17/3

Second, we need to multiply 10 by a fraction that will give us the denominator of 3.

10 * 3/3 = (10*3)/3 = 30/3

Third, we do subtraction using our derived fractions.

30/3 - 17/3 = (30-17)/3 = 13/3

Lastly, we simplify the improper fraction. Improper fraction is a fraction whose numerator is greater than its denominator. Its simplified form is a mixed fraction.

13/3 = 4 1/3

Arliss needs to buy 4 1/3 yards more to complete the required yard length.

6 0

3 years ago

Read 2 more answers

Jan plans to spend $100 on a pair of jeans and shirts. She buys 2 shirts for $24 each. What is the most she can spend on the jea

Volgvan

Answer:

Step-by-step explanation:

24+24=48

100-48=52

6 0

3 years ago

Read 2 more answers

Find the equation of the line that passes through he points (2,6) and (-2,4)

stepan [7]

First Find slope by using $\frac{y2-y1}{x2-x1}$

$\frac{4-6}{-2-2} = \frac{-2}{-4} = \frac{1}{2}$

Now use point Slope form to find the equation of the line (y-y₁)=m(x-x₁)
(y-6)= $\frac{1}{2}$ (x-2)
y-6= $\frac{1}{2} x-1$
y= $\frac{1}{2} x+5$ is your equation

7 0

2 years ago

30 points 1 question PLEASE HELP ME OUT I REALLY NEED IT

Pachacha [2.7K]

Answer:

p(-5/3) ≠ 0 So, (3 x +5) is NOT A FACTOR of p(x)

Step-by-step explanation:

Here, the given function is $p(x)=3x^5+2x^2 - 5$

Now, the given root of the function is ( 3x +5)

Now, if ( 3 x + 5) = 0,

we get x = - 5/3

So, the zero of the given polynomial is x = -5/3

Then, x = -5/3, p(x) =0 ⇒ ( 3 x + 5) is a FACTOR of p(x)

Now, let us find the value of function at x = -5/3

Substitute x = -5/3 in the given function p(x), we get:

$p(x)=3x^5+2x^2 - 5 \implies p(\frac{-5}{3}) = 3(\frac{-5}{3})^5 + 2(\frac{-5}{3})^2 - 5\\= 3(\frac{-3,125}{243}) + 2(\frac{25}{9}) - 5\\= (\frac{-3,125}{81}) + (\frac{50}{9}) - 5\\= -38.580 + 5.56 - 5 = -38.02\\\implies p(\frac{-5}{3}) = -38.02$

Now, as p(-5/3) ≠ 0 So, (3x +5) is NOT A FACTOR of p(x)

6 0

3 years ago

ANSWER ALL 5 PARTS.

N76 [4]

A function $f$ from a set A to a set B is defined as a relation that assings to each element $x$ in the set A exactly one element $y$ in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is $x$ feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let $x$ be the width of the tomato patch, then the function that matches this statement is:

$\boxed{Wa(x)=\frac{x}{2}}$

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

 $\boxed{La(x)=x+2}$

2. Area of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then the area of a rectangle is the product of the length and width. So:

 $A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}$

3. Combined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

 $Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}$

4. Write the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:

 $Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x$

5. Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

 $31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06$

Therefore the area of the remaining space is:

 $\boxed{Aa(7.06)=7.06ft^{2}}$

6 0

3 years ago