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3 years ago

I will mark brainliest if correct please answer fast

Mathematics

2 answers:

Shtirlitz [24]3 years ago

8 0

Answer: 264 cm^2

Step-by-step explanation:

22*6.5+22*5.5=22*12=264

Anon25 [30]3 years ago

5 0

The area of the given figure 156.75 sq cm.

Step-by-step explanation:

Here,

The figure is a trapezium.

In this trapezium

a = 22 cm

b = 6.5 cm

A and b are the length of two parallel sides.

Height (h) = 5.5 cm

To find the area of the figure which is consists of two trapezium

Formula

Area of the trapezium = $\frac{1}{2} (a+b)h$

Now,

The area of one trapezium = $\frac{1}{2} (22+6.5)X5.5$ sq cm

= 78.375 sq cm

So,

The area of the figure = 78.375×2 sq cm = 156.75 sq cm

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the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?

madam [21]

Answer:

About 0.6548 grams will be remaining.

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$

5 0

2 years ago

Find the oth term of the geometric sequence 7, 14, 28, ...

yaroslaw [1]

Answer:

The nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

$a_n=a_1\cdot r^{n-1}$

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

$\frac{14}{7}=2,\:\quad \frac{28}{14}=2$

The ratio of all the adjacent terms is the same and equal to

$r=2$

now substituting r = 2 and a₁ = 7 in the nth term

$a_n=a_1\cdot r^{n-1}$

$a_n=7\cdot \:2^{n-1}$

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

$a_n=7\cdot \:2^{n-1}$

6 0

2 years ago

Margaret has 16 yards of fabric and she is making dresses for z doll write the expression for the length of fabric used for each

emmasim [6.3K]

<h3>The expression for the length of fabric used for each doll is $\frac{16}{z}\ yards$ </h3>

<em><u>Solution:</u></em>

Given that,

Margaret has 16 yards of fabric

she is making dresses for z doll

Therefore,

Total length of fabric = 16 yards

Number of dresses = z dolls

Therefore,

$\text{ length of fabric used for each doll } = \frac{ \text{ Total length of fabric }} {\text{ z dolls }}\\\\\text{ length of fabric used for each doll } = \frac{16}{z}\ yards$

Thus, the expression for the length of fabric used for each doll is $\frac{16}{z}\ yards$

7 0

3 years ago

The power of ten representation of four thousandths

m_a_m_a [10]

Answer:

if its thousandths, its 4 x 10^-1. If you mean 4000, its 4 x 10^3

Step-by-step explanation:

5 0

2 years ago

I need a models or visuals for that problem

Leya [2.2K]

Answer:I will be a mode

Step-by-step explanation:

4 0

3 years ago