All categories

3 years ago

Please Answer this ill brainliest

Mathematics

1 answer:

daser333 [38]3 years ago

8 0

Answer:

30 onions in each row. I think he could because 600/10 is 60.

You might be interested in

Thaddeus models the number of hours of daylight in his town as

AfilCa [17]

Option (A) : least: 10 hours; greatest: 14 hours

The function f(x) = sin x has all real numbers in its domain, but its range is

−1 ≤ sin x ≤ 1.

How to solve such range questions?

Such questions in which every term is in addition and its range is asked is simplest ones to solve if we know the range of each of term. This can be seen from this question

Given: d(t) = 2sin(xt) + 12

= −1 ≤ sin (xt) ≤ 1.

= −2≤ 2 sin (xt) ≤ 2.

= 10 ≤ 2sin (xt) + 12 ≤ 14

= 10 ≤d(t) ≤ 14

Thus least: 10 hours; greatest: 14 hours

Learn more about range of trigonometric ratios here :

brainly.com/question/14304883

#SPJ4

6 0

3 years ago

Write the amount of water in each glass in milliliters.there are 1/20 liters

Liono4ka [1.6K]

Answer:

There are 50 milliliters, Hope this helps. try to figure out why on your own!

Step-by-step explanation:

5 0

3 years ago

if a sailboats main sail is shaped like a right triangle and the base is 18m long. if it makes an angle of 60°, as marked, how t

mestny [16]

Check the picture below.

8 0

3 years ago

Becky drew drawing of a sign before creating it. Her scale drawing had a width of 6 inches, and the actual sign has a width of 4

Vlad1618 [11]

Answer:

3 in. = 2ft

Step-by-step explanation:

6 in. = 4ft

3 in. = 2ft

5 0

3 years ago

I need to know q2 please help

Phantasy [73]

Given that the half-life of uranium is $4.5\times10^9$ years, you know that after that amount of time you're left with half as much uranium as what you start with:

$\dfrac12=e^{-k\times4.5\times10^9}$

Solve this equation for $k$ to find the decay rate.

$\dfrac12=e^{-k\times4.5\times10^9}$
$\implies \ln\dfrac12=\ln e^{-k\times4.5\times10^9}$
$\implies -\ln2=-k\times4.5\times10^9\ln e$
$\implies k=\dfrac{\ln2}{4.5}\times10^{-9}$

Now you're looking to find the time it takes for the substance to decay to one eighth of its original amount, which is the time $t$ such that

$\dfrac18=e^{-kt}$
$\implies\ln\dfrac18=\ln e^{-kt}$
$\implies-\ln8=-kt\ln e$
$\implies t=\dfrac{\ln8}k=\dfrac{\ln8}{\frac{\ln2}{4.5}\times10^{-9}}=\dfrac{\log_28}{4.5}\times10^9\approx0.66\times10^9$

5 0

4 years ago