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3 years ago

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On Monday, the high temperature was 60°. On Tuesday, the high temperature was 12° colder than on Monday. On Wednesday, the high

temperature was 10° warmer than the temperature on Tuesday. What was the temperature on Wednesday, in degrees Fahrenheit? Show your work or explain your answer.

1 answer:

prisoha [69]3 years ago

3 0

Answer:

68 degrees

Step-by-step explanation:

60-12(12 degrees colder on Tuesday)=58

58+10(10 degrees warmer on Wednesday)=68

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The equation of parabola is y = 3x²- 12x+ 15.

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

The equation of the parabola is given as y = x².

If it has a vertex then the equation can be written as,

y = a(x-h)² + k

where (h, k ) is the vertex and a is the stretching or compressing factor.

So here the vertex is (2,3) and it is vertically stretched by 3.

So the equation is given as,

y = 3(x-2)² + 3

y = 3(x² -4x+4) + 3

= 3x² - 12x + 12 + 3

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Step-by-step explanation:

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3 years ago

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Solve by elimination <br> 7x+7y=-7<br> -10x-7y=28

Zarrin [17]

Answer:

x = - 7, y = 6

Step-by-step explanation:

7x + 7y = - 7 -----> equation 1

- 10x - 7y = 28 -----> equation 2

Add equations 1 & 2,

7x + 7y = - 7

- 10x - 7y = 28

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- 3x = 21

x = 21 / - 3

x = - 7

Substitute x = - 7 in equation 1,

7x + 7y = - 7

7y + 7x = - 7

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3 years ago

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On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)

xz_007 [3.2K]

Answer:

$P = 10 + 2\sqrt{53}$ units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3), X (2, 3),

Y (4, -4), Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

For WX:

$(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)$

$WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}$

$WX = \sqrt{(-5)^2 + (0)^2}$

$WX = \sqrt{25}$

$WX = 5$

For XY:

$(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)$

$XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}$

$XY = \sqrt{-2^2 + (3 +4)^2}$

$XY = \sqrt{-2^2 + 7^2}$

$XY = \sqrt{4 + 49}$

$XY = \sqrt{53}$

For YZ:

$(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)$

$YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}$

$YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}$

$YZ = \sqrt{7^2 + (-2)^2}$

$YZ = \sqrt{49 + 4}$

$YZ = \sqrt{53}$

For ZW:

$(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)$

$ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}$

$ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}$

$ZW = \sqrt{0^2 + (-5)^2}$

$ZW = \sqrt{0 + 25}$

$ZW = \sqrt{25}$

$ZW = 5$

The Perimeter (P) is as follows:

$P = WX + XY + YZ + ZW$

$P = 5 + \sqrt{53} + \sqrt{53} + 5$

$P = 5 + 5 + \sqrt{53} + \sqrt{53}$

$P = 10 + 2\sqrt{53}$ units

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4 years ago