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3 years ago

11

Round 569.903507815 to the nearest whole number?

1 answer:

Monica [59]3 years ago

3 0

Answer:

if you round it to a whole number... it should be 570

Step-by-step explanation:

just round the tenth place

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Given the midpoint (1.5,1.5) and the endpoint (5,7) where is the other endpoint located

earnstyle [38]

The formula of a midpoint:

$M_{AB}\left(\dfrac{x_A+x_B}{2},\ \dfrac{y_A+y_B}{2}\right)$

We have:

$M(1.5,\ 1.5)\to x_M=1.5,\ y_M=1.5\\A(5,\ 7)\to x_A=5,\ y_A=7$

Substitute

$\dfrac{5+x_B}{2}=1.5\qquad|\cdot2\\\\5+x_B=3\qquad|-5\\\\x_B=-2\\\\\dfrac{7+y_B}{2}=1.5\qquad|\cdot2\\\\7+y_B=3\qquad|-7\\\\y_B=-4$

<h3>Answer: (-2, -4)</h3>

8 0

3 years ago

Can someone help me and you need to round to nearest whole number

andrew11 [14]

Answer:

look at the picture i have sent

5 0

2 years ago

What is the 10th term of 6,10,14,18?

Tasya [4]

Y = 4x + 2
y = 4×10 + 2
y = 42

the 10th term is 42

4 0

3 years ago

Read 2 more answers

Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p

kvv77 [185]

The question is an illustration of composite functions.

Functions c(n) and h(n) are $\mathbf{c(n) = 5000 + 250n}$ and $\mathbf{n(h) = 5h}$
The composite function c(n(h)) is $\mathbf{c(n(h)) = 5000 + 1250h}$
The value of c(n(100)) is $\mathbf{c(n(100)) = 130000}$
The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

$5000 in fixed costs plus an additional $250
5 systems in one hour of production

<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

$\mathbf{c(n) = Fixed + Additional \times n}$

This gives

$\mathbf{c(n) = 5000 + 250 \times n}$

$\mathbf{c(n) = 5000 + 250n}$

The function for number of systems is:

$\mathbf{n(h) = 5 \times h}$

$\mathbf{n(h) = 5h}$

<u>(b) Function c(n(h))</u>

In (a), we have:

$\mathbf{c(n) = 5000 + 250n}$

$\mathbf{n(h) = 5h}$

Substitute n(h) for n in $\mathbf{c(n) = 5000 + 250n}$

$\mathbf{c(n(h)) = 5000 + 250n(h)}$

Substitute $\mathbf{n(h) = 5h}$

$\mathbf{c(n(h)) = 5000 + 250 \times 5h}$

$\mathbf{c(n(h)) = 5000 + 1250h}$

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

$\mathbf{c(n(100)) = 5000 + 1250 \times 100}$

$\mathbf{c(n(100)) = 5000 + 125000}$

$\mathbf{c(n(100)) = 130000}$

<u>(d) Interpret (c)</u>

In (c), we have: $\mathbf{c(n(100)) = 130000}$

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

5 0

3 years ago

1. ADEA has vertices D(8,4), E(2, 6), and F(3, 1). What are the vertices of the image after a dilation with a scale factor of 5

lubasha [3.4K]

Answer:

The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

Step-by-step explanation:

If the point (x, y) dilated by a scale factor k using the origin as a center of dilation, then its image is (kx, ky)

Let us use this rule to solve our question

∵ The vertices of ΔDEF are D (8, 4), E (2, 6), and F (3, 1)

∵ ΔDEF dilated by a scale factor 5

∴ k = 5

∵ The origin as the center of dilation

→ That means, multiply the coordinates of every point by 5

∴ D' = (8 × 5, 4 × 5)

∴ D' = (40, 20)

∴ E' = (2 × 5, 6 × 5)

∴ E' = (10, 30)

∴ F' = (3 × 5, 1 × 5)

∴ D' = (15, 5)

∴ The vertices of the image are D' (40, 20), E' (10, 30), F' (15, 5)

3 0

2 years ago