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sweet-ann [11.9K]

3 years ago

13

Triangle ABC has vertices A(−4, 3) , B(2, 3) , and C(−5, 1) . A dilation with a scale factor of 5 and center at the origin is ap

plied to this triangle.
What are the coordinates of C′C′ in the dilated image?

C′ has a coordinate pair of (
,
)

1 answer:

inna [77]3 years ago

5 0

The coordinates of C'C' is (-25, 5) and all you have to do is multiply each point by 5.

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A rectangular carpet has a perimeter of 274 inches. The length of the carpet is 95 inches more than the width. What are the dime

GrogVix [38]

Answer:

Length: 116

Width: 21

Step-by-step explanation:

Let's say that the width of carpet is x. If the length of the carpet is 95 inches more than the width, then let's say that the width is x + 95.

Since we are solving for the perimeter, you have to multiply the width and the length(x and x + 95) by two.

So,

2x + 2x + 190 = 274.

Simplify this and you get:

4x= 274 - 190

4x = 84

x= 21

So, the width is 21. This means that the length is 21 + 95.

21 + 95 = 116

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Y=<br> Pls help solve for

miss Akunina [59]

Answer:

$y=39\degree$

Step-by-step explanation:

The given triangle is a right angle triangle .

The known sides of the triangle that is opposite to angle y is 8 units while the side adjacent is 10 units.

The value of y can be calculated using the tangent ratio.

$\tan(y)=\frac{8}{10}$

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3 years ago

Which statement represents the given algebraic expression?

alexandr1967 [171]

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2 years ago

The sum of five consecutive multiples of 4 is 20.Find the smallest integer.

Aloiza [94]

Answer:

The smallest integer is -4.

Step-by-step explanation:

Let the smallest integer be x.

Since the integers are all multiples of 4, they are

=> x, x+4, x+8, x+12 and x+16

=> x + x+4 + x+8 + x+12 + x+16 = 20

=> 5x + 40 = 20

Subtract 40 from both sides of the equation

5x +40 - 40 = 20 - 40

5x = -20

Divide both sides by the coefficient of x(which is 5)

$\frac{5x}{5} = \frac{-20}{5}$

x = -4

∴ the smallest integer is -4.

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2 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

2 years ago