All categories

3 years ago

5th grade math. correct answer will be marked brainliest

Mathematics

2 answers:

Otrada [13]3 years ago

8 0

Answer:

Six times

Step-by-step explanation:

Because it is Six times six over seven, therefore there are "6 shaded models"

juin [17]3 years ago

3 0

X 6

i’m pretty sure this is right

You might be interested in

Please help need ASAP

andrey2020 [161]

Octagon is 309.02cm²
Pentagon is 172.05in²

7 0

2 years ago

Read 2 more answers

PLEASE ANSWERING THIS QUESTION!!!

Anarel [89]

Answer:

104.8cm³(1 decimal place)

Step-by-step explanation:

Volume of a cone is given by the formula

V=1/3πr²h

we are given a radius of 5 and a height of 4 so we will just substitute for the values and we will give our pie as 22/7

V=1/3× 22/7×5²×4

V=1/3 × 22/7 ×25 ×4

V=2200/21

V=104.76190476

V=104.8cm³(1decimal place)

pls if you like this answer you can appreciate it by marking it as brainliest

4 0

8 months ago

Write the slope-intercept form of the equation of the line through the given points through {-3,1} and (-5,5)

Zigmanuir [339]

Slope = -2/1

work is provided in the image attached.

3 0

3 years ago

Please help ...........

sammy [17]

Answer:

Angle B = 80°

see attachment......

7 0

2 years ago

Element X decays radioactively with a half life of 12 minutes. If there are 200 grams of Element X, how long, to the nearest ten

Semenov [28]

Answer:

It would take 24 minutes for the element to decay to 50 grams

Step-by-step explanation:

The equation for the amount of the element present, after t minutes, is:

$Q(t) = Q(0)e^{-rt}$

In which Q(X) decays radioactively with a half life of 12 minutes.(0) is the initial amount and r is the rate it decreases.

Half life of 12 minutes

This means that $Q(12) = 0.5Q(0)$

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-12r}$

$e^{-12r} = 0.5$

$\ln{e^{-12r}} = \ln{0.5}$

$-12r = \ln{0.5}$

$12r = -\ln{0.5}$

$r = -\frac{\ln{0.5}}{12}$

$r = 0.05776$

If there are 200 grams of Element X, how long, to the nearest tenth of a minute, would it take the element to decay to 50 grams?

This is t when Q(t) = 50. Q(0) = 200.

$Q(t) = Q(0)e^{-rt}$

$50 = 200e^{-0.05776t}$

$e^{-0.05776t} = 0.25$

$\ln{e^{-0.05776t}} = \ln{0.25}$

$-0.05776t = \ln{0.25}$

$0.05776t = -\ln{0.25}$

$t = -\frac{\ln{0.25}}{0.05776}$

$t = 24$

It would take 24 minutes for the element to decay to 50 grams

6 0

3 years ago