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3 years ago

Please help.worth 40 point

Mathematics

2 answers:

malfutka [58]3 years ago

7 0

Answer:

64 is a perfect square that provides 8 rows , and each contains 8 seats, this is the first perfect square after 50

Step-by-step explanation:

Lilit [14]3 years ago

3 0

64 is a perfect square that provides 8 rows , and each contains 8 seats, this is the first perfect square after 50

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I need help with my algebra 2. I also want to know how to solve this.

Lisa [10]

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

$x+1 \geq 0 \iff x \geq -1$

Under this assumption, we square both sides and we have

$x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0$

The solutions to this equation are

$x = -2,\ x=1$

But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$ .

In fact, we have

$x=-2 \implies \sqrt{-2+3}-1=0\neq -2$

and

$x=1 \implies \sqrt{4}-1=1$

which is the only solution.

4 0

3 years ago

It takes Prakash 2 1/2 hours to paint a medium sized room. If he can work 50 minutes each day, how many days will it take to com

Aleksandr-060686 [28]

60 min in one hour so 60+60+30=150.
150 divided 50 mins = 3 days (50+50+50=150)

3 0

3 years ago

Help me with this question plzzzz

zubka84 [21]

V = L * W * H
V = 1/4 * 2/3 * 3/5
V = (1 * 2 * 3) / (4 * 3 * 5)
V = 6/60 reduces to 1/10 m^3 <==

5 0

3 years ago

Find an equation for the plane through the points p=(0,0,0), q=(0,1,0), and r=(1,2,3)

jek_recluse [69]

Hello,

Assume Ax+By+Cz+D=0 the plane 's equation.

p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0

q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0

r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3

An equation of the plane is -3x+z=0

6 0

3 years ago

Complete the statements.

spayn [35]

Answer:

1. 30

2. 150

Step-by-step explanation:

$3tan^2(x)-1 =0$

Lets assume tan(x) = u

$3u^2(x)-1 =0$

Now we solve for 'u'

add 1 on both sides

$3u^2(x) =1$ , divide both sides by 3

$u^2 = \frac{1}{3}$

Take square root on both sides

$u = +-\frac{1}{\sqrt{3} }$

We replace tan(x) for 'u'

$tan(x) = +-\frac{1}{\sqrt{3} }$

x = 30 because $tan(30) =+\frac{1}{\sqrt{3} }$ in first quadrant

x = 30 (tan is positive in first quadrant)

$tan(x) =-\frac{1}{\sqrt{3} }$

x = 150 because $tan(150) =-\frac{1}{\sqrt{3} }$ in second quadrant

tan is negative in second quadrant

4 0

3 years ago