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Pavlova-9 [17]
3 years ago
13

Good morning temperature is -15° but the afternoon the temperature raises by 20° right in addition expressions for the afternoon

temperature vent to shoulders and find the temperature in the afternoon
Mathematics
1 answer:
stealth61 [152]3 years ago
7 0

Answer:

+ 5°

Step-by-step explanation:

Given,

Good morning temperature = -15°

Raise in temperature = 20°

Afternoon temperature = -15° + 20°

           = + 5°

Hence, temperature in after noon is equal to + 5°.

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Simplify expressions
charle [14.2K]
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<h2>-  \frac{2}{5}  \div  \frac{3}{4}</h2><h3>■To divide by a fraction, multiply by the reciprocal of that fraction</h3>

<h2>-  \frac{2}{5}  \times  \frac{4}{3}</h2><h3>■Multiply the fractions</h3>

<h2>-  \frac{8}{15}</h2>

<h3>Hence, Quotient =-  \frac{8}{15}</h3>

<h3>b) 0.4 \div  ( - 0.75)</h3><h3>■Convert the decimals into a fractions</h3>

<h2>\frac{2}{5}  \div ( -  \frac{3}{4} )</h2><h3>■Dividing a positive and a negative equals a negative: (+)÷(-)=(-)</h3>

<h2>-  \frac{2}{5}  \div  \frac{3}{4}</h2><h3>■To divide by a fraction, multiply by the reciprocal of that fraction</h3>

<h2>-  \frac{2}{5}  \times  \frac{4}{3}</h2><h3>■Multiply the fractions</h3>

<h2>-  \frac{8}{15}</h2><h3>Hence, Quotient is -  \frac{8}{15}</h3>

<h3>c)-  \frac{2}{5}  \div  \frac{3}{4}</h3><h3>■To divide by a fraction, multiply by the reciprocal of that fraction</h3>

<h2>-  \frac{2}{5}  \times  \frac{4}{3}</h2><h3>■Multiply the fractions</h3>

<h2>-  \frac{8}{15}</h2><h3>Hence, The Quotient is -  \frac{8}{15}</h3>
4 0
3 years ago
Calculate the perimeter of this shape.
Sloan [31]

Step-by-step explanation:

Hello there!

Just double the given sides!

(18*2)+(12*2)=60 cm.

There you are!

:)

4 0
3 years ago
Find an equation of the line passing through the pair of points. Write the equation in the form Ax +By = C.
Alexxandr [17]

9514 1404 393

Answer:

  5x -y = -37

Step-by-step explanation:

One way to find the coefficients A and B is to use the differences of the x- and y-coordinates:

  A = Δy = y2 -y1 = 2 -(-3) = 5

  B = -Δx = -(x2 -x1) = -(-7 -(-8)) = -1

Then the constant C can be found using either point.

  5x -y = 5(-7) -2 = -37

The equation of the line is ...

  5x -y = -37

_____

<em>Additional comment</em>

This approach comes from the fact that the slope of a line is the same everywhere.

  \dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\Delta y}{\Delta x}\\\\\Delta x(y-y_1)=\Delta y(x-x_1)\qquad\text{cross multiply}\\\\\Delta y(x-x_1)-\Delta x(y-y_1)=0\qquad\text{subtract y term}\\\\\Delta y(x) -\Delta x(y) = \Delta y(x_1)-\Delta x(y_1)\qquad\text{Ax+By=C form}

The "standard form" requires that A be positive, so we chose point 1 and point 2 to make sure that was the case.

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Answer:

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