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3 years ago

Write the equation (use any form you choose) of the line that passes through the points (4, 3) and (-2, 6)

1 answer:

5 0

Gradient
$m = \frac{y - y}{x - x} \\ m = \frac{6 - 3}{ - 2 - 4} \\ m = \frac{ - 1}{2}$
$equation \\ y = mx + c \\ 3 = \frac{ - 1}{2} \times(4) + c \\ 3 = - 2 + c \\ c =5$

THE ANSWER IS :-
$y = mx + c \\ y = - \frac{1}{2} x + 5$

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Work out the area of the shaded region.

Makovka662 [10]

Answer: Area of the shaded region is 6.867 square cm

Step-by-step explanation:

The shaded area corresponds to the difference of the area of the rectangle and the area of the semicircle.

Area of the rectangle = 4 times 8 = 32 cm^2

Area of the semicircle = 0.5 * pi * r^2 (where r is the radius, i.e., half of the diameters, i.e., 4 cm). So Area = 0.5*pi*4^2 = 25.133 cm^2

The difference: 32 - 25.133 = 6.867 cm^2 (area of the shaded region)

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3 years ago

Which correctly describes the type of symmetry of the butterfly pictured?

Alexxandr [17]

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3 years ago

-4(X + 3) s-2 - 2x What is the solution?

steposvetlana [31]

The answer would be -6x +s -14

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3 years ago

Please help i really need to get a good grade

solong [7]

Replace f(x) with x⁸

1) f(x) + 2 → x⁸ + 2

2) 3f(x) → 3x⁸

3) f(-x) → (-x)⁸

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3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$