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3 years ago

Solve for p: √p − 1 − √p = 1

Mathematics

1 answer:

astraxan [27]3 years ago

4 0

√p − 1 − √p = 1
- 1 = 1 (√p is canceled out)
since -1 cant equal 1 so there's no solution

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Graph x^3 - 3x^2 + 2 = 0.1 - x What are the solutions of the equation?

liq [111]

Answer:

$x=-0.6,x=1.52,x=2.08$

Step-by-step explanation:

Let $f(x)=x^3-3x^2+2$ .

and

$g(x)=0.1-x$

The graph of the two equations are shown in the attachment.

The points where the two graphs intersected are: $(-0.6,0.7),(1.52,-1.42),(2.08,-1.98)$

The x-coordinates of the intersection points are the solutions to $x^3-3x^2+2=0.1-x$ .

Therefore the solutions are: $x=-0.6,x=1.52,x=2.08$

6 0

2 years ago

If the numbers below were ordered from least to greatest, which number could you use to replace the blank? 0.20,2/5 □ , 3/4,8/10

faust18 [17]

1/2 would work for the blank space

because as shown in the image you could relate it to a whole for each one and put it to 100 to figure out what the blank could be

the 0.20 can be converted to 1/5 so that way you can put everything into the 100 group

8 0

3 years ago

Determine the next step for solving the quadratic equation by completing the square.

Studentka2010 [4]

Answer:

The solutions are $x=\frac{1+\sqrt{7}}{2}$ and $x=\frac{1-\sqrt{7}}{2}$

Step-by-step explanation:

we have

$0=-2x^{2}+2x+3$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$-3=-2x^{2}+2x$

Factor the leading coefficient

$-3=-2(x^{2}-x)$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$-3-0.5=-2(x^{2}-x+0.5^{2})$

$-3.5=-2(x^{2}-x+0.5^{2})$

Rewrite as perfect squares

$-3.5=-2(x-0.5)^{2}$

$7/4=(x-0.5)^{2}$

square root both sides

$x-0.5=(+/-)\sqrt{\frac{7}{4}}$

$x-\frac{1}{2}=(+/-)\frac{\sqrt{7}}{2}$

$x=\frac{1}{2}(+/-)\frac{\sqrt{7}}{2}$

$x=\frac{1+\sqrt{7}}{2}$

$x=\frac{1-\sqrt{7}}{2}$

4 0

3 years ago

50 points! Solve: 5a + 6.9 = 3.6a +16.28

icang [17]

A = 6.7 :) hope this helps I double checked

6 0

2 years ago

Read 2 more answers

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

3 years ago