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2 years ago

Question 2 of 5

Mathematics

1 answer:

Ipatiy [6.2K]2 years ago

5 0

It may be B—JL is longer than JK

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Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

3 0

6 months ago

D Evaluate arcsin (6)] at x = 4. dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

2 years ago

Add the coefficients 2x + 4x - 8x - 5x *

Ber [7]

2+4-8-5=9

=-7x

Add the coefficients

3 0

2 years ago

Read 2 more answers

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term.

Simora [160]

Consecutive terms in an arithmetic sequence differ by a constant $d$ . So

$5k-1=2k+d$
$6k+2=5k-1+d$
$\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5$

Denote the $n$ -th term in the sequence by $a_n$ . Now that $a_1=2$ , we have

$a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5$

which means

$a_8=2+(8-1)\cdot5=37$

4 0

2 years ago

Write the sum as a product. Simplify the product.

postnew [5]

Answer:

Step-by-step explanation:

We see (-10) repeating six times. This means the product would be 6(-10). To simplify this, 6 × (-10) = -60.

3 0

2 years ago

Read 2 more answers