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professor190 [17]
3 years ago
14

Need help with 6 ASAP

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
8 0
The answer to that question is d
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Tia had a garden that was 2 meters long. Her brother used 458 millimeters of the length at one end as a worm farm
Nastasia [14]

Answer:

154 centimeters

Step-by-step explanation:

Each meter is 1000 millimeters or 100 centimeters, and if Tia's brother used 458 millimeters of 2000 millimeters, you can subtract 458 from 200 and you get 1542. Since the answer options are in centimeters, simply divide by 10 and you get the answer in centimeter form (154)

6 0
4 years ago
Read 2 more answers
Which is bigger 3/8 or 0.3
Luba_88 [7]
\frac{3}{8} \\
0.3=\frac{3}{10}

If two fractions have the same numerator, the fraction with the smaller denominator is bigger.
8\ \textless \ 10 \Rightarrow \frac{3}{8} \ \textgreater \  \frac{3}{10}

3/8 is bigger.
3 0
3 years ago
Read 2 more answers
darcy likes to eat peanut butter and raisins on apple slices. on each apple slice she puts 1/16 cup of peanut butter and 8 raisi
netineya [11]
The answer is 7/10.

For one apple slice, Darcy needs 1/16 cup of peanut butter and 8 raisins. She has 2/5 cup of peanut butter. To compare how many butter she has and she needs we will divide 2/5 with 1/1

This means that Darcy needs 5/80 cup of peanut butter <span>and 8 raisins.
If she has 32/80 cup of peanut butter, she will have it only fo 6 or 7 apple slices (32/80 </span>÷ 5/80 = 6.4) depending how it is rounded. If she prepares 6 apple slices, she will still have a little of peanut butter left (0.4 = 2/5 of cup). So, she will prepare 7 apple slices so the butter will be all gone.

For 7 apple slices, she needs 56 raisins (7 × 8 raisins).
That is 56 raisins out of 80 raisins, so <span>fraction of the 80 raisins did she eat is:
</span><span>\frac{56}{80} = \frac{7}{10}</span>
4 0
3 years ago
4(3h+7)/4+h for h=-2
fomenos
The answer would be 2
8 0
3 years ago
A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between
sattari [20]

Answer:

a)

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) 0.09

Step-by-step explanation:

We are given the following in the question:

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)

b) rate of increase after one day.

We put t = 1

B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09

The rate of increase after 1 day is 0.09

8 0
3 years ago
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