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2 years ago

Edmund makes a cube using eight small cubes.

Mathematics

1 answer:

anyanavicka [17]2 years ago

8 0

Answer: 184

Step-by-step explanation:

so edmund would have a 2 by 2 cube (try and draw it out if that helps) so

2 x 2 = 4

2 x 3 = 6

2 x 4 = 8

so samuel cubeboid would be

long = 4

wide = 6

hight = 8

8 x 6 x 4 = 192 small cubes

but we are trying to find how much more samuel got

so 192 - 8 = 184

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8. If3:8=c:56, the value of c will be

adelina 88 [10]

The scale for the second values is 56/8 = 7

C = 3 x 7 = 21

C= 21

6 0

2 years ago

The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by

motikmotik

Answer:

t = 25.1seconds

Step-by-step explanation:

X(t)= 16cos75°t

Y(t)= 6 + (Vosin75°t - 16t^2)

Converting 78 yards to feet :

1 yard = 3 feet

78feet =?

78×3=234 feet

Y(t)= number of feet above the ground at t seconds

VoCos75° t = 234

t = 234/(VoCos75°)

At this time,4 = y(234/VoCos75°)

4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2

Vo= 140.66ft/s

Time,t= 234/(140.66× Cos75°)

t = 25.087seconds

t= 25.1 seconds (1 decimal place)

8 0

2 years ago

Beth solved the system of equations by making a substitution for y in the second equation. What equation did she get as a result

Slav-nsk [51]

Hello there!

The correct answer is option C

3x + 20x - 55 = -9

I am not sure if you wanted some more details but I hope this helps and if you do, just let me and I will edit it.

Good luck with your studies!

Happy New Year!~

3 0

2 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 $[\log3,\infty)$ .