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ELEN [110]
3 years ago
8

The record high in January temperature in Austin,Texas,is 90 degreasEr. The record low January temperature is-2 degressF. Find t

he difference between the high and low temperatures
Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0
Difference just means you subtract the two numbers, so:

90 - (-2) = x

( now you would "Copy Change Opposite") so;

90 + 2 = 92 *F


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Answer: -3x+7

Step-by-step explanation: Good luck! :D

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Please help!!!!!
WARRIOR [948]

3)

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{-7}~,~\stackrel{y_2}{7}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -7 -2}{2}~~~ ,~~~ \cfrac{ 7 + 9}{2} \right)\implies \left( \cfrac{-9}{2}~~,~~\cfrac{16}{2} \right)\implies \left( -4\frac{1}{2}~~,~~8 \right)

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3 years ago
Please help! .........
stira [4]

Answer:

1) When x is 1, y is 4, and vice versa

2) When x is -6, y is 42, and when x is 7, y is 114

3) When x is -2, y is 2, and when x is 8, y is 42

Step-by-step explanation:

1)

y=x^2-6x+9

y+x=5 which can be rearranged as y=5-x.

Substituting this lone y into the first equation, you get:

5-x=x^2-6x+9

Move everything to one side:

x^2-5x+4=0

Factor:

(x-4)(x-1)=0

x=1, 4

y=4, 1

2)

y-30=12x which can be rearranged as y=12x+30.

y=x^2+11x-12

Let's use elimination this time and subtract the first equation from the second:

0=x^2-x-42

Factor:

(x-7)(x+6)=0

x=-6, 7

y= -42, 114

3)

y=x^2-2x+6

y=4x+10

Let's set the two equations equal to each other through substitution:

x^2-2x-6=4x+10

Move everything to one side:

x^2-6x-16=0

Factor:

(x-8)(x+2)=0

x=-2, 8

y= 2, 42

Hope this helps!

6 0
3 years ago
Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.
Darina [25.2K]

Answer:

\cos( \theta)  =  \frac{5}{8}  \\  {8}^{2}  =  {5}^{2}  +  {opp}^{2}  \\ {opp}^{2}  = 64 - 25 = 39 \\ opp =  \sqrt{39}  \\  \sin(\theta)  =  \frac{\sqrt{39}}{8}  \\  \tan(\theta)  =  \frac{ \sqrt{39} }{5}  \\  \cot(\theta)  =  \frac{1}{\tan(\theta)}  =  \frac{1}{\frac{ \sqrt{39} }{5}}  \\ \cot(\theta)  =  \frac{5}{ \sqrt{39} }  \\  \csc( \theta) =  \frac{1}{\sin(\theta)}  =  \frac{1}{\frac{\sqrt{39}}{8}}   \\ \csc( \theta) = \frac{8}{ \sqrt{39} }  \\  \sec( \theta) =  \frac{1}{\cos( \theta) }  =  \frac{1}{ \frac{5}{8} }  \\ \sec( \theta) =  \frac{8}{5}

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2 years ago
Arnold needs to simplify the expression below.
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Answer:

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3 0
3 years ago
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