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melamori03 [73]

3 years ago

10

The temperature is falling outside at a steady rate of 4 degrees Fahrenheit every hour. If the temperature starts at 68 Fahrenhe

it write a linear equation that relates the Fahrenheit temperature, F, to the time in hours, t, that it has been falling. Function notation is F(h)=.

1 answer:

Aleksandr-060686 [28]3 years ago

4 0

Answer:

F(t)= -4t+68

Thats all I can say :)

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What is the slope of a line parallel to the line with equation 5x + 3y = 7?

weeeeeb [17]

Answer:

-5/3

Step-by-step explanation:

Parallel lines are lines which have the exact same slope but different y-intercepts. We can find the slope by converting to slope-intercept form, y=mx+b from standard form.

We convert by using inverse operations to isolate y.

5x+3y=7

5x-5x+3y=7-5x

0x+3y=-5x+7

3y=-5x+7

$\frac{3y}{3} =\frac{-5x+7}{3} \\y=\frac{-5}{3}x+\frac{7}{3}$ .

The slope is -5/3. SInce parallel lines have the same slope, the slope for a parallel line will be -5/3.

3 0

3 years ago

Find ABC please someone answer quick !!!!

ankoles [38]

Answer:

Same for the question user's profile

ANSWER: 118DEGREES

Step-by-step explanation:

3x+11=4x-5

x=6

3(6)+11

29

29*2

59

59*2=118

6 0

2 years ago

What is the range of the function in the graph below(photo imported)

olga55 [171]

Answer: y is less than or equal to -1

How?

The graphs highest value it will ever reach, range wise, is -1. Therefore it isnt 0, or all real numbers.

The reason why it is -1 and not -2 is because the curve hits -1. Sure, -2 is included in the range, but y is less than or equal to -1 INCLUDES -2

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6 0

3 years ago

If quadrilateral ABCD is congruent to quadrilateral EFGH, then what corresponding side is congruent to side FG? A) Side AB

BartSMP [9]

Two congruent quadrilaterals have congruent corresponding sides and angles. This means that if quadrilateral ABCD is congruent to quadrilateral EFGH, then:

AB≅EF;
BC≅FG;
CD≅GH;
AD≅EH;
∠A≅∠E;
∠B≅∠F;
∠C≅∠G;
∠D≅∠H.

Answer: correct choice is B (FG is congruent to side BC)

4 0

3 years ago

5. During the eruption of Mount St. Helens in 1980. debris was ejected at a speed of over 440 feet per second

skelet666 [1.2K]

Answer:

3500 ft

Step-by-step explanation:

We assume you intend the equations to be ...

$y(t)=-16t^2+425t+8200\\x(t)=113t$

Then we can solve the second equation for t and use that in the first equation. We can solve the first equation for y=6000.

x = 113t

t = x/113

y(t) = y(x/113) = -16(x/113)² +425(x/113) +8200 = 6000

(16/12769)x² -(425/113)x -2200 = 0 . . . . . . subtract 6000, multiply by -1

Using the quadratic formula, the positive solution is ...

x = ((425/113) +√((-425/113)² -4(16/12769)(-2200))/(2(16/12769))

The fraction in the denominator can be inverted and used as a multiplier. We also use a√b = √(a²b) to simplify the radical.

x = (48025 +√4104275825)/32

x ≈ 3502.8

Rock was thrown about 3500 feet horizontally.

7 0

3 years ago