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3 years ago

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On Thursday,the afternoon temperature was 5c lower than the morning temperature.The temperature at night was 3c and it was 8c hi

gher than the afternoon temperature.
What is the morning, afternoon,and night temperature?

2 answers:

Mumz [18]3 years ago

8 0

Answer:

The answer is:

morning: 0c

afternoon: -5c

night: 3c

Step-by-step explanation:

Given that night temperature is 3c

and night temperature was 8c higher than the afternoon temperature

so 3c-8c = -5c

afternoon temperature was 5c lower than the morning temperature

so -5c +5c = 0c

Vaselesa [24]3 years ago

4 0

Night = 3degree

Afternoon = 8degrees less than night (3degrees)
--> 3 – 8 = –5degrees

Morning = –5 Dregree more than the afternoon
--> –5 + 5 = 0degree

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3 years ago

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jeka57 [31]

Answer: 13.18

Step-by-step explanation:

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3 years ago

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A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses

mash [69]

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for $x$ days.

So, for $x$ days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = $\frac{360}{x}$

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = $(x+4)$ days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= $(\frac{360}{x}-3)$

Total expense in dollars can now be given as: $(x+4)(\frac{360}{x}-3)$

Simplifying by using distribution (FOIL).

$(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))$

$360-3x+\frac{1440}{x}-12$

$348-3x+\frac{1440}{x}$

We know total expense remains the same which is = $360.

So, we have the equation as:

$348-3x+\frac{1440}{x}=360$

Multiplying each term with $x$ to remove fractions.

$348x-3x^2+1440=360x$

Subtracting $348x$ both sides

$348x-348x-3x^2+1440=360x-348x$

$-3x^2+1440=12x$

Dividing each term with -3.

$\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}$

$x^2-480=-4x$

Adding $4x$ both sides.

$x^2+4x-480=-4x+4x$

$x^2+4x-480=0$

Solving using quadratic formula.

For a quadratic equation: $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in values from the equation we got.

$x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}$

$x=\frac{-4\pm\sqrt{16+1920}}{2}$

$x=\frac{-4\pm\sqrt{1936}}{2}$

$x=\frac{-4\pm44}{2}$

So, we have

$x=\frac{-4+44}{2}$ and $x=\frac{-4-44}{2}$

$x=\frac{40}{2}$ and $x=\frac{-48}{2}$

∴ $x=20$ and $x=-24$

Since number of days cannot be negative, so we take $x=20$ as the solution for the equation.

Thus, the original duration of the tour = 20 days

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bija089 [108]

Hey There!

Lets Just Work through your answers,

ⓧ A - An isosceles triangle is a triangle with (at least) two equal sides, so this would not apply to the given triangle.

:) B - A scalene triangle is a triangle in which all three sides have different lengths so this applies to the given triangle. One side measures 10, one measures 11, and one measures 12.1

ⓧ C - A right triangle is a type of triangle that has one angle that measures 90 degrees, which this triangle does not so this does not apply.

ⓧ D - In geometry, an equilateral triangle is a triangle in which all three sides are equal, so this does not apply to the given triangle.

ⓧ E - An obtuse triangle is a triangle with one obtuse angle (greater than 90°) and two acute angles, so this does not apply to the given triangle.

:) F - An acute triangle (or acute-angled triangle) is a triangle with three acute angles less than 90 degrees. So this applies to the given triangle.

In Summary, B & F classify the given triangle correctly.

If you could rate five starts & give brainliest that would be greatly appreciated!

Hope I helped, Have a good day.

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3 years ago