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2 years ago

Suppose a candy bar weighs 5.47 ounces. How much does a pack of 6 bars weight

Mathematics

2 answers:

olchik [2.2K]2 years ago

8 0

Answer:

32.82 OUNCES

Step-by-step explanation:

It is really easy.

All you have to do is 5.47x6

Nina [5.8K]2 years ago

3 0

Answer:

32.82

Step-by-step explanation:

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Which best describes the solution set of the compound inequality below?

Olenka [21]

Answer:

All real numbers except x=1.

8 0

3 years ago

If the measure of an angle is 13 degree, find measure 0f its supplement

amid [387]

Answer:

167 degrees

Step-by-step explanation:

supplementary angles equal to be 180 degrees, so subtract the given angle from 180 to find the missing angle.

180-13=167

5 0

3 years ago

Please answer correctly !!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!

vaieri [72.5K]

Add the ratios together:

5 + 2 + 1 = 8

Divide the length of NQ by that:

24/8 = 3

Multiply the ratio of NO by that/l

The ratio of NO is 5, 5 x 3 = 15

NO = 15

3 0

3 years ago

MEC makes a big family tent in the shape of a symmetrical triangular prism. The floor of the tent is 5 m long and 2.5 m wide. Th

SVEN [57.7K]

Answer:

15,236 sq in

Step-by-step explanation:

First, you have to split the net into three rectangles and two triangles

step 1

triangular base area

1/2 bh = 1/2 (44 in)(64 in)

= 1/2 (2,816 sq in)

= 1,408 sq in.

2 x 1,408 sq in = 2,816 sq in.

step 2

end rectangle area

lw = (68 in)(69 in)

= 4,692 sq in

2 x 4,692 sq in = 9,384 sq in

Now find the area of the middle rectangle

lw - (69 in) (44 in) = 3,039 sq in.

then add all the areas together

2,816 + 9,384 + 3,036 = 15,236 sq in

Hope this helps :)

3 0

3 years ago

Prove the theorem (AB )^T= B^T. A^T

Lisa [10]

Answer:

$(AB)^T = B^T.A^T (Proved)$

Step-by-step explanation:

Given (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

$(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\ = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T} \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)$

3 0

3 years ago