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4 years ago

12

In a city, the record monthly high temperature for March is 56°F. The record monthly low temperature for March is -4°F. What is

the range of temperatures for the month of March?

1 answer:

katrin2010 [14]4 years ago

8 0

The range can be expressed as t1 - t2

t1 = 56

t2 = - 4 (That's Cold)

Range = 56 - - 4

Range = 56 + 4 = 60o F

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We have two points, (3,0) and (0,-3). So we can find the slope of the line:

m=(y2-y1)/(x2-x1)

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y=x-3 subtract x from both sides

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3 years ago

Tamiami turned in 6 homework assignments. if she turned in 75 percent of the homework assignments that were due how many homewor

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3 years ago

Suppose Cotangent (theta) = StartFraction 5 Over 12 EndFraction, where Pi less-than theta less-than StartFraction 3 pi Over 2 En

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Answer:

Sin theta = 12/13

Step-by-step explanation:

From the question;

Cot theta = 5/12

Kindly recall;

Cot theta = 1/ tan theta

Hence, tan theta = 12/5

Mathematically

tan theta= opposite/adjacent

to get hypotenuse, we will use Pythagoras’ theorem which states that the square of the hypotenuse equals sum of the squares of the two other sides

let hypotenuse be h

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h^2 = 169

h = √169

h = 13

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3 years ago

2 u + 12 < 23<br> 0,1 d + 8 > 0<br> 3 - 4 r > 7<br> 13 < 2 z + 3<br> 6 ≥ 9 - 0, 15 i

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Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

<u><em>The question is</em></u>

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

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The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

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3 years ago

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3 years ago