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3 years ago

If you put a sugar cube in hot tea does it dissolve or melt and why?

2 answers:

8 0

If you put a sugar cube in hot tea it will dissolve because the heat from the tea will make the sugar cube deform and the sugar from the sugar would mix/dissolve or sink to the bottom of the tea.

iren2701 [21]3 years ago

3 0

The molecules in the sugar dissolved in water. it did not dissociate though, just dissolved. the atoms in each individual molecule stayed in tact, but the intermolecular forces broke apart.

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1. What is the area of the shape? 6cm 4cm 2cm 3cm

Solnce55 [7]

Hello,
I’m not sure how to answer your question because looks like you forgot to put an image of the shape your talking about.

3 0

3 years ago

Ani spent a total of $7.53 at a store. She bought some cereal for $3.67. She also bought 6 peaches.

Ratling [72]

Answer:

Step-by-step explanation:

It says she spent $7.53 at the store and she bought (SOME) cereal for $3.67

so we have to find how much the 6 peaches cost. To find it out we are going to have to subtract the cereal she bought from $3.67 and the money she wasted total so its going to be D.

8 0

2 years ago

Help asap for brainlist help asap

Mariana [72]

Answer:

2x^3+11x^2-43x-24

Step-by-step explanation:

desmos graphing calculator- both factors matched

8 0

2 years ago

Please help ASAP! Thanks! For the figures below, assume they are made of semicircles, quarter circles and squares. For each shap

bazaltina [42]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)$

$A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)$

$C=6\pi\ cm$

The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)

$AC=\sqrt{12^2+12^2}$

$AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

6 0

2 years ago

Someone please help me?

vekshin1

Coordinates of the center you can find from
x-3=0 and y+5=0
x=3 and y=-5
(3,-5) is the center of the circle.

25 is a square of the radius.
r²=25, so
r=5

Center (3,-5), r=5
So, the answer is D.

4 0

3 years ago