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Maurinko [17]
3 years ago
15

Please help ^-^ 1. 23 = x – 9.5 2.12 = k – 4.3

Mathematics
1 answer:
Alborosie3 years ago
3 0

Answer:

1. 32.5

2. 16.3

Step-by-step explanation:

This is algebra. We need to find the unknown values.

1. 23 = x – 9.5

  Adding 9.5 on both sides of the equation

  23 + 9.5 = x – 9.5 + 9.5

  32.5 = x

∴ x = 32.5

2. 12 = k – 4.3

Adding 4.3 on both sides of the equation

12 + 4.3 = k – 4.3 + 4.3

16.3 = k

∴ k = 16.3

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Maru [420]

Answer:

3

Step-by-step explanation:

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3 years ago
What is the equation of this trend line?<br><br> Enter your answers by filling in the boxes.
zhuklara [117]

Answer:

k=-2j+24

Step-by-step explanation:

Take two points from the graph

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m=\frac{12-24}{6-0}

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we have

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Note: The y-intercept is given (0,24)

substitute

k=-2j+24

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4 years ago
Use the net to find the lateral area of the cylinder.
ioda
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Read 2 more answers
Point B is 2/3 the way from A(-2,5)to C(4,-4). Find the location of B
mr Goodwill [35]

well, we'll first off put the point AC in component form by simply doing a subtraction of C - A, multiply that by the fraction 2/3, and that result will get added to point A, to get point B.

\bf \textit{internal division of a segment using a fraction}\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-4})~\hfill \frac{2}{3}\textit{ of the way from }A\to C \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_2}{4}-\stackrel{x_1}{(-2)}, \stackrel{y_2}{-4}-\stackrel{y_1}{5})\implies (4+2,-9) \stackrel{\textit{component form of segment AC}}{\qquad \implies \qquad (6,-9)} \\\\[-0.35em] ~\dotfill

\bf \stackrel{~\hfill \textit{coordinates to be added to point A}} {\cfrac{2}{3}(6,-9)\implies \cfrac{2}{3}(6)~,~\cfrac{2}{3}(-9)\qquad \implies \qquad \left(4,-6\right)} \\\\\\ \stackrel{\textit{additions to point A}} {\stackrel{\textit{point A}}{(-2,5)}+\left( 4,-6\right)}\implies \left( -2+4~~,~~5-6\right) = B\implies (2~,~-1)=B

5 0
3 years ago
Part 1: Linear Systems and Line Segments
Nataly_w [17]

Answer:

y=8(x−8)(x+0.5)

Step-by-step explanation:

Part 1: Linear Systems and Line Segments

Create and attach a 1 page ‘cheat’ sheet for the Linear Systems Unit.

Solve each system by graphing:

y = x+2, y = 4-1

2x – y + 9 = 0, x + y – 3 = 0

Solve by substitution:

x – 2y = 4, 2x – 3y = 7

Solve by elimination:

3x + 8y = -3, x + 8y = -5

4x – 5y = -22, 5x +6y = -3

Find the length of the line segment joining each pair of points. (Express to nearest 10th of a unit)

(2,5) and (5,7)

(-1, -3) and (-2, -4)

Find the midpoint of each line segment.

M (-2, 2), P (4, 8)

A (3, 3), J (-9, 3)

The Midpoint of AB is M(-2, 1). IF A(-7, 3), what are the coordinates of B?

Page Break

Part 2: Trigonometry

Create and attach a 1 page ‘cheat’ sheet for the Trigonometry Unit.

Solve each right triangle:

Shape E

Shape F 62⁰ D

150 cm

ShapeText BoxA

Shape15⁰

ShapeB C

Page Break

Solve each triangle:

Shape L

58⁰

56⁰

M 19 cm N

ShapeJ

58⁰

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K L

Khalid wants to be sure his boat is safely anchored. He knows that the angle of depression that the boat makes with the horizontal when anchored should be less than 12⁰. The boat is 100 m above the seabed and the anchor cable is 440 m long. Is Khalid safely anchored? Explain.

Part 3: Polynomials

Create and attach a 1 page ‘cheat’ sheet for the Polynomial Unit.

Simplify

(4x - 3) + (5x+4)

(2x2 – 3x + 4) – (x2 + 4x - 1)

(-6x3 yz2 )(-2xy2 z)

Factor, if possible

7x + 42

X2 + 7x + 12

3x2 = 12x = 9

X2 + 9

5x2 – 20x + 20

Page Break

Part 4: Quadratics

Create and attach a 1 page ‘cheat’ sheet for the Quadratics Unit.

Graph the data and write the relation in the form of

y=a(x−h)2

y=ax−h2

X

Y

-6

32

-5

18

-4

8

-3

2

-2

0

-1

2

Graph the data and write the relation in the form of

y=a(x−h)2

y=ax−h2

X

Y

-1

-8

0

-4.5

1

-2

2

-0.5

3

0

4

-0.5

Graph the relation by plotting the vertex and 2 other points.

y=(x+1)2−1

y=x+12−1

y=−2(x−6)2+2

y=−2x−62+2

For each, write the relation in standard form:

y=ax2+bx+c

y=ax2+bx+c

a=2, b=5, y intercept = 11

y=5x2+bx+c

y=5x2+bx+c

, vertex at (1, 4)

a=−3

a=−3

, passes through (2, 11) and (0, -5)

Factor:

x2−17x+66

x2−17x+66

x2−3x−18

x2−3x−18

x2+13x

x2+13x

x2−9

x2−9

Find the zeros.

y=(x+3)(x−3)

y=x+3x−3

y=−10x2

y=−10x2

y=8(x−8)(x+0.5)

y=8(x−8)(x+0.5)

4 0
3 years ago
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