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2 years ago

Ella went trick-or-treating with her family. When

Mathematics

2 answers:

Aleksandr [31]2 years ago

6 0

Answer:

6.6 because she put her candy bag on a scale and if each piece of candy weighs 12 grams,

Kazeer [188]2 years ago

5 0

Answer:

Step-by-step explanation:

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Using the triangle below, answer the following questions.

fomenos

Step-by-step explanation:

Answer(1):

Law of Cosines.

Answer(2):

since side "c" is missing so we will write formula used for side "c"

$c^2=a^2+b^2-2ab\cdot\cos\left(C\right)$

Answer(3):

First lets write both sine and cosine formulas:

Check the attached picture for the list of formulas:

From given picture we see that two angles A and B are missing. Also 1 side "c" is missing.

Sine formula uses two angles while cosine formula uses only one angles.

Hence cosine formula will be best choice to find the missing values.

4 0

3 years ago

Whats the gcf for 10y5+30y3-15y

Nataly [62]

Answer:

1, see the picture below, I used a calculator!

I hope this helps!!

5 0

3 years ago

How to multiply mixed fractions

morpeh [17]

Answer:

Step-by-step explanation:

Change each number to an improper fraction.

Simplify if possible.

Multiply the numerators and then the denominators.

Put answer in lowest terms.

Check to be sure the answer makes sense.

8 0

3 years ago

4) Find the total cost of a $0.95 pen with a 60% markup.

weqwewe [10]

$1.52 is the total cost. Good luck and have an amazing day!

7 0

3 years ago

Suppose each edge of the cube shown in the figure is 10 inches long. Find the sine and cosine of the angle formed by diagonals D

tatiyna

Check the picture below.

$sin(EDG )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{10\sqrt{3}}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{sin(EDG )=\cfrac{\sqrt{3}}{\sqrt{3^2}}\implies sin(EDG )=\cfrac{\sqrt{3}}{3}} \\\\[-0.35em] ~\dotfill$

$cos(EDG )=\cfrac{\stackrel{adjacent}{10\sqrt{2}}}{\underset{hypotenuse}{10\sqrt{3}}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{cos(EDG )=\cfrac{\sqrt{6}}{\sqrt{3^2}}\implies cos(EDG )=\cfrac{\sqrt{6}}{3}}$

4 0

2 years ago