All categories

3 years ago

PLZ HELP WILL GIVE BRAINLY IF CORRECT

Mathematics

2 answers:

MAXImum [283]3 years ago

8 0

Answer:

Step-by-step explanation:

1,000mm=100cm

pochemuha3 years ago

3 0

Answer:

1000 MM is the greatest

Step-by-step explanation:

1000 MM to CM = 100

9.7 CM and 4.7 CM are too small, therefore, we ignore them..

450 MM to CM is 45, therefore 1000 MM is the greater length

You might be interested in

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force F⃗ =( 35 N )i^−( 37 N )j^ to

Trava [24]

Answer:

-189.8 J

Step-by-step explanation:

We are given that f

Force applied=F=35 Ni-37 Nj

Displacement, s=(-8.7m)i-(3.1m)j

We have to find the work done .

We know that

Work done= $F\cdot s$

Using the formula

Work done= $(35i-37j)\cdot (-8.7-3.1)$

Work done= $-304.5+114.7$

Using $i\cdot i=j\cdot j=k\cdot k=1,i\cdot j=j\cdot i=j\cdot k=k\cdot j=i\cdot k=k\cdot i=0$

Work done=-189.8J

Hence, the work done =-189.8 J

5 0

3 years ago

The MoviePlex theater sold 1,456 tickets. The theater sold 6 times as many regular price tickets as it sold discount tickets. Ma

Lady bird [3.3K]

Jonah is the one that is correct because 1/6th of 1248 is 208. Then you add 1248 and 208 which is 1456 which is the correct answers

5 0

3 years ago

What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .