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4 years ago

6

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 t

he range of temperature for the months of March

1 answer:

guapka [62]4 years ago

7 0

If I am correct the range is just the space in between the high and the low so you just at add the two together 56+4=60 degree Fahrenheit

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rhombus and a square have one and the same side of 6 cm. The area of the rhombus is 4/5 of the area of the square. Find the heig

siniylev [52]

Consider this option:
1. area_rombus=a*h, where a=6 - the length of the side, h - height.
h=area_rombus/a.
2. area_sq=a², where a=6 - the length of the square.
area_sq=36, area_rombus=4/5 *36=28.8.
3. according to the item 1 h=area_rombus/a=28.8/6=4.8.

answer: 4.8

6 0

4 years ago

PLEASE HELP THANK YOU SO MUCH

FinnZ [79.3K]

As you can probably tell, this is a 30, 60, 90 triangle. That means the degree measures are 30, 60, and 90. These are special triangles. For every 30, 60, 90 triangle in the world, the shorter leg (in your case, x) is ALWAYS two times shorter than the hypotenuse (6). 6/2=3, so x=3. Also, the longer leg for a 30, 60, 90 triangle is always equal to the shorter leg times the square root of 3. So, 3 times the square root of 3 is about 5.2. So, y=5.2. Hope it helps!

4 0

3 years ago

Which expression shows the result of applying the distributive property to 14(−3n+32)14(−3n+32) ?

strojnjashka [21]

The Distributive Property is an algebra property which is used to multiply a single term and two or more terms inside a set of parentheses.so the resulting expression when you of apply the distributive property to 14(−3n+32)= (14) (-3n) + (14) (32)= -42n + 448

8 0

3 years ago

a goods train leaves a station at 6 pm followed by an express train which leaves at 8 pm and travels 20 km/hour faster than the

kolbaska11 [484]

Answers:

Goods train has a speed of 80 km per hour

Express train has a speed of 100 km per hour

==========================================================

Explanation:

Check out the diagram below to see the table. For now, refer to table 1.

Each row represents a different train.

We have x as the speed of the slower train (the goods train) and x+20 is the speed of the faster train (express train). These speeds are in km/hr or kph.

The y variable represents the number of hours it takes the goods train to arrive at its destination. The equation formed in the first row is 1040 = xy. This is because distance = rate*time.

Let's solve for y to get

1040 = xy

xy = 1040

y = 1040/x

We'll use this equation later.

-------------------------

The second row of table 1 has x+20 as the speed and y-2-0.6 as the time.

The "-2" represents the fact that the express train is operating for 2 hours less compared to the goods train (because the goods train had a 2 hour head start). Then the "-0.6" indicates we're taking off another 36 minutes, which is equivalent to 36/60 = 0.6 hours.

So overall, the express train is traveling for 2+0.6 = 2.6 fewer hours compared to the goods train. The express train's time value is y-2.6 hours.

With this in mind, we can form this equation for the second row

distance = rate*time

1040 = (x+20)(y-2.6)

We'll now apply substitution to replace y with 1040/x. This is valid because earlier we found that y = 1040/x

------------------------

If we apply that substitution and solve for x, we get the following

$1040 = \left(x+20\right)\left(y-2.6\right)\\\\1040 = \left(x+20\right)\left(\frac{1040}{x}-2.6\right)\\\\1040 = x\left(\frac{1040}{x}-2.6\right) + 20\left(\frac{1040}{x}-2.6\right)\\\\1040 = 1040 - 2.6x + \frac{20800}{x} - 52\\\\0 = -2.6x + \frac{20800}{x} - 52\\\\x*0 = x*\left(-2.6x + \frac{20800}{x} - 52\right)\\\\0 = -2.6x^2 + 20800 - 52x\\\\-2.6x^2 - 52x + 20800 = 0\\\\$

Let's now use the quadratic formula to finish things off.

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-52)\pm\sqrt{(-52)^2-4(-2.6)(20800)}}{2(-2.6)}\\\\x = \frac{52\pm\sqrt{219024}}{-5.2}\\\\x = \frac{52\pm468}{-5.2}\\\\x = \frac{52+468}{-5.2} \ \text{ or } \ x = \frac{52-468}{-5.2}\\\\x = \frac{520}{-5.2} \ \text{ or } \ x = \frac{-416}{-5.2}\\\\x = -100 \ \text{ or } \ x = 80\\\\$

Ignore the negative solution because we can't have a negative speed.

The only practical solution here is x = 80

This means the goods train has a speed of 80 km/hr.

The express train's speed must be x+20 = 80+20 = 100 km/hr.

------------------------

Now to verify these answers.

If the goods train travels 1040 km and does so at a speed of 80 km/hr, then it travels for 1040/80 = 13 hours. Add this onto 6:00 PM and we arrive at 7:00 AM. This is shown in the first row of table 2 (see below).

If the express train travels 1040 km and its speed is 100 kph, then it travels for 1040/100 = 10.4 hours. This is equivalent to 10 hours, 24 minutes because 0.4 hrs = 0.4*60 = 24 min

If we start at 8:00 PM and elapse 10 hours, then we'll arrive at 6:00 AM. Add on another 24 minutes, and we get to 6:24 AM. Notice how this is 36 minutes before 7:00 AM (because 24+36 = 60). So this confirms that the express train indeed arrives 36 minutes before the goods train does, despite the goods train having that 2 hour head start.

5 0

3 years ago

HELP URGENT PLZ PLZ PLZ

Sindrei [870]

Answer: D. $y=\frac{1}{9} x^{2}$

Step-by-step explanation: The smaller the coefficient of a fraction by the quadratic function (the x squared value), the more narrow the graph becomes. 1/4 is larger than 1/9 and the other two whole numbers are still larger than 1/9. Even though -1 is less than 1/9, this only makes the graph flip, not make it any more narrow. Fractions are what squeeze graphs, negatives do not make a difference (i.e. -1/4 is the same size of narrowness as 1/4).

6 0

3 years ago