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Ket [755]
2 years ago
12

Can someone please help I pay £100

Mathematics
1 answer:
Flura [38]2 years ago
3 0

Answer:

The new points to the triangle will be:
A(1,4)\\ B(1,1)\\ C(-1,1)

Step-by-step explanation:

Because the reflection point is at x=2, all x values will subtract their distances from x=2 to get their new values. The y values remain the same.

The starting values are:

A(3,4)\\ B(3,1)\\ C(5 ,1)

Point A is 1 unit away from x=2, so we'll subtract 1 from 2 to get the new x value: 2 - 1 = 1, so A(1,4).

 

Point B is also 1 unit away from x=2, so we'll subtract 1 from 2 to get the new x value: 2 - 1 = 1, so B(1,1).

Point C is 3 units away from x=2, so we'll subtract 3 from 2 to get the new x value: 2 - 3 = -1, so C(-1,1).

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Which expression is equivalent to 24?<br> O 2√3<br> 23√/3<br> O 2√6<br> O 23√√6
vredina [299]

24^{1/3}=\sqrt[3]{24}=\sqrt[3]{8 \cdot 3}=\sqrt[3]{8}\sqrt[3]{3}=\boxed{2\sqrt[3]{3}}

5 0
1 year ago
PLEASE HELP ASAP!! Kate is trying to convert an area from meters squared to centimeters squared. She multiplied the area she had
Savatey [412]

Answer:

1m=100cm

1m squared=100cm×100cm

1msquared= 10000cm squared

5 0
1 year ago
I'm have some issues with the problem shown in the screenshot and would love some help.
KiRa [710]

The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;

  • Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.

  • Maximum volume of the box is approximately 1048.6 in.³

<h3>How can the dimensions and volume of the box be calculated?</h3>

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let <em>x </em>represent the side lengths of the cut squares, we have;

Width of the box formed = 18 - 2•x

Length of the box = 35 - 2•x

Height of the box = x

Volume, <em>V</em>, of the box is therefore;

V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x

By differentiation, at the extreme locations, we have;

\frac{d V }{dx}  =  \frac{d( 4 \cdot \:  {x}^{3}  - 106 \cdot \:  {x}^{2}  + 630\cdot \:  {x} )  }{dx} = 0

Which gives;

\frac{d V }{dx}  =12\cdot \:  {x}^{2}  -  212\cdot \:  {x} + 630 = 0

6•x² - 106•x + 315 = 0

x  =  \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}

Therefore;

x ≈ 4.55, or x ≈ -5.55

When x ≈ 4.55, we have;

V = 4•x³ - 106•x² + 630•x

Which gives;

V ≈ 1048.6

When x ≈ -5.55, we have;

V ≈ -7450.8

The dimensions of the box that gives the maximum volume are therefore;

  • Width ≈ 18 - 2×4.55 in. = 8.89 in.

  • Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.

  • Height = x ≈ 4.55 in.

  • The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

#SPJ1

7 0
1 year ago
The number of cars sold weekly by a new automobile dealership grows according to a linear growth model. The first week the deale
marusya05 [52]

The number of cars that sold on the third week is (P3=26)

The number of cars that sold on the first week is (P4=33)

<u>Step-by-step explanation:</u>

<u>Given:</u>

  • The number of cars that sold on the first week is (P0=7)
  • The number of cars that sold on the second week is (P0=12)

We have to find the number of cars being sold on the upcoming week

From the data given above, frame the equation

Pn = Pn −1+7   ( 12-5=7  it denotes the cars sold in the first and the second  week)

Pn=5+7n (cars in the first week and the cars sold in the second week into "n" n is used to find the cars sold in the upcoming weeks)

(If n=3)

Pn=5+7(3)

Pn=26

The number of cars that sold on the third week is (P3=26)

(If n=4)

Pn=5+7(4)

Pn=33

The number of cars that sold on the first week is (P4=33)

5 0
3 years ago
PLEASE Help...
Alex787 [66]
Like in money?Is there a specific amount?. or anything more to what I would have?I need more details
4 0
3 years ago
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