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3 years ago

Rectangle EFGH is graphed on a coordinate grid. EFGH is transformed according to the rule , resulting in rectangle E’F’G’H’.

Mathematics

2 answers:

Alona [7]3 years ago

6 0

im sorry i dont understand that brother like thats sum weird

marshall27 [118]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is the difference between these fractions?<br><br> +3/4−(+7/8)

Ket [755]

Answer:

1/8

Step-by-step explanation:

3/4 is 6/8 and 6/8 - 7/8 is 1/8

4 0

2 years ago

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Find the distance CD rounded to the nearest tenth<br> C=(10,-1) and D=(-6,3)

Fynjy0 [20]

Answer:

CD = 16.5

Step-by-step explanation:

To find the distance between two points, use this formula: $L = \sqrt{(x_{2} -x_{1})^{2}+(y_{2} -y_{1})^{2} }$

point C can be info set 1: (10, -1) x₁ = 10 y₁ = -1

point D can be info set 2: (-6, 3) x₂ = -6 y₂ = 3

Substitute the information into the formula

$L = \sqrt{(x_{2} -x_{1})^{2}+(y_{2} -y_{1})^{2} }$

$CD = \sqrt{(-6 -10)^{2}+(3 -(-1))^{2} }$ Simplify inside each bracket

$CD = \sqrt{(-16)^{2}+(4)^{2} }$ Square the numbers

$CD = \sqrt{256+16 }$ Add inside the root

$CD = \sqrt{272}$ Enter into calculator

$CD = 16.5$ Rounded to the nearest tenth, the first decimal

The distance CD is 16.5.

4 0

4 years ago

Choose the equation that represents this situation. Use n to represent the

Flauer [41]

Answer:

Step-by-step explanation:

I know is C let em have brainliest

5 0

3 years ago

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The number N of cars produced at a certain factory in 1 day after t hours of operation is given by Upper N (t )equals 800 t minu

Elden [556K]

Answer:

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

Step-by-step explanation:

The function N(t) = 800 t - 5t², represents the number of cars produced at a time t hours in a day, where 0 ≤ t ≤ 10

The function C(N) = 30,000 + 8,000 N, represents the cost C (in dollars) of producing N cars

We need to find The cost C as a function of the time t

That means Substitute N in C by its function by other word find the composite function (C о N)(t)

∵ C(N) = 30,000 + 8,000 N

∵ N(t) = 800 t - 5 t²

- Substitute N in C by 800 t - 5 t²

∴ C(N(t)) = 30,000 + 8000(800 t - 5 t²)

- Multiply the bracket by 8000

∴ C(N(t)) = 30,000 + 8000(800 t) - 8000(5 t²)

∴ C(N(t)) = 30,000 + 6,400,000 t - 40,000 t²

- C(N(t) = C(t)

∴ C(t) = 30,000 + 6,400,000 t - 40,000 t²

The cost C as a function of t is C(t) = 30,000 + 6,400,000 t - 40,000 t²

5 0

3 years ago

What is 4 5/13 as a decimal rounded to the nearest tenth

MAXImum [283]

Answer:

4.4

Step-by-step explanation:

4 5/13 = 4 + 5/13 = 4 + 0.384615... = 4.384615...

Answer: 4.4

5 0

3 years ago

Read 2 more answers