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Alex787 [66]
3 years ago
10

Between the time Iko woke up and lunchtime, the temperature rose by 11°. Then by the time he went to bed, the temperature droppe

d by 14°.
Write an addition expression for the temperature relative to when Iko woke up.
Mathematics
1 answer:
Tanzania [10]3 years ago
4 0

Answer:             -3 degrees

Step-by-step explanation:

Lets say that the original temperature was "x."

The temperature rose by 11 degrees so it is now at 11+x.

Then the temperature fell by 14 degrees so you subtract. 11+x-14

11-14= -3 so you subtract 3 degrees from the original temperature and there fore the relative temperature is -3

I hope this helps!!!

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Veronika [31]

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So, for the given function to be defined, we need to find the possible values for which the values of x makes the square root to be positive.

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-9 -5x ≥ 0

Now, let's solve for x

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8 0
1 year ago
given the points a (-1,2) and (7,14), find the coordinates of the point p on directed line segment ab that partitions ab in the
Vinvika [58]
Check the picture below.

\bf \left. \qquad  \right.\textit{internal division of a line segment}
\\\\\\
a(-1,2)\qquad b(7,14)\qquad
\qquad 1:3
\\\\\\
\cfrac{aP}{Pb} = \cfrac{1}{3}\implies \cfrac{a}{b} = \cfrac{1}{3}\implies 3a=1b\implies 3(-1,2)=1(7,14)
\\\\
-------------------------------\\\\
{ P=\left(\cfrac{\textit{sum of "x" values}}{r1+r2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{r1+r2}\right)}

\bf -------------------------------\\\\
P=\left(\cfrac{(3\cdot -1)+(1\cdot 7)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 14)}{1+3}\right)
\\\\\\
P=\left( \cfrac{-3+7}{4}~,~\cfrac{6+14}{4} \right)

3 0
3 years ago
1. Evaluate -3x^3 -2y^2 when x=-5 and y =3
-Dominant- [34]

Answer:

357

Step-by-step explanation:

A) We replace x = - 5 and u = 3 into -3x^3 -2y^2

and we have -3x^3 -2y^2= -3*(-5)^3 -2*3^2= -3*(-125)- 2*9

or -3x^3 -2y^2 = 375-18= 357

The answer is 357.

B) Replace h = -2 into h^2-3h+2

we have h^2-3h+2 = (-2)^2 -3*(-2) +2 = 4 +6 + 2= 12

The answer is 12

Have a good day.

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