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3 years ago

What the answer to the question

Mathematics

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

Step-by-step explanation:

When we know two sides and the included angle, there is a formula we can use.

We know angle C = 35º, and sides a = 20cm and b = 19.5cm.

using the formula

Area =(½)ab sin C(angle)

Put in the values we know:

(20×19.5÷2)×sin(35)=

111.8474050885 cm^2

rounded = 111.85 cm^2

answer is c

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What is the area, in square inches, of a circle with a diameter equal to 16 inches?

Lostsunrise [7]

Answer: 201.06 in²

Step-by-step explanation:

Use the equation πr² to find out the area of a circle.

Since the diameter is 16 the radius is half of that, 8.

π(8)²

Work out that equation and you're done!

7 0

3 years ago

Which points lies on the line described by the equation? y+4=(x-3)

NikAS [45]

Hello :
the equation <span>y+4=(x-3) for the line passes by the point : (3 , - 4 )
when the slope is : 1 </span>

3 0

3 years ago

What ratio is proportional to 18/21

Doss [256]

Answer:

There are many.

Step-by-step explanation:

When you say proportional, you are looking for a ratio that is equivalent to the ratio given. In your case, there are many so you might need to be more specific. But still, we can help you figure it out.

An easy way to do this would be to scale it down to its simplest form and then move upwards. To find proportional ratios, just multiply denominator and the denominator with the same factor.

$\dfrac{18}{21} = \dfrac{6}{7}$

That is your ratio in its simplest form. Now we can scale up, I'll show you how to do one completely:

$\dfrac{6}{7} x \dfrac{2}{2} = \dfrac{12}{14}$

That's the same ratio scaled up by a factor of 2.

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$\dfrac{6}{7} x \dfrac{3}{3} = \dfrac{18}{21}$

That's the ratio given

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$\dfrac{6}{7} x \dfrac{4}{4} = \dfrac{24}{28}$

Scaled up by a factor of 4

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$\dfrac{6}{7} x \dfrac{5}{5} = \dfrac{30}{35}$

Scaled up by a factor of 5.

The list goes on and on.

3 0

3 years ago

The actual Mount Rushmore carving was made from a scale model with a scale of 1 inch = 1 foot. On the model, Teddy Roosevelt's m

IRISSAK [1]

Answer:

Step-by-step explanation:

The carving was made from a scale model with a scale of 1 inch = 1 foot. This means that one foot on the actual carving is represented by one inch on the model. So the model is smaller than the actual carving.

On the model, Teddy Roosevelt's mustache was 1 foot by 8 inches long.

We would convert the 1 foot on the model to inches because the model is represented in inches

12 inches = 1 foot

This means that on the model, Teddy Roosevelt's mustache was 12 inches by 8 inches long. Therefore,

Teddy Roosevelt's mustache was 12 feets by 8 feets long on the monument or carving

8 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

3 years ago