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4 years ago

PLEASE HELP 3 MORE!! images below

Mathematics

2 answers:

Harman [31]4 years ago

4 0

Answer:

The Corect answer would be B

Step-by-step explanation:

If you slice a triangler prism horizontally, and look at it from the top, it will look like a 2D triangle

stealth61 [152]4 years ago

3 0

Answer:

Figure B

Step-by-step explanation:

In order to get 3 of the triangles if the shape were to be cut horizontally, you need a shape with a triangle base, and a prism.

In this case, figure B contains both of those requirements.

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Which of the following is a correct way to name this angle? B, 2 ACB А, САВ D. BCA C. Z CBA

olga nikolaevna [1]

The answer is Angle ACB

The angle is form from both line A and C

3 0

1 year ago

Help Me On One Of These Please!

gayaneshka [121]

We need more information on the stuff before we can answer it

8 0

3 years ago

2- log 6-log 9<br> In condensing

Tju [1.3M]

Answer:

$\log(150)$

Step-by-step explanation:

We want to rewrite the following as a single logarithm:

$2 - log(6) - log(9)$

We introduce logarithm in the first term to get:

$2 \log(10) - \log(6) - \log(9)$

Recall that:

$log(a) - log(b) = \log \frac{a}{b}$

We apply this property to get:

$2 log(10) - log( \frac{6}{9} )$

This gives:

$log( {10}^{2} ) - log( \frac{6}{9} )$

$log( {100}) - log( \frac{2}{3} )$

$log( \frac{100}{ \frac{2}{3} } ) = log( \frac{300}{2} ) = log(150)$

4 0

3 years ago

If you want four 2 in. platys, seven 1 in. guppies, and a 3 in. loach, what is the smallest capacity tank you can buy?

nikklg [1K]

Answer:

168 cubic inches

Step-by-step explanation:

The parameters given are:

four 2 in. platys, seven 1 in. guppies, and a 3 in. loach,

Four 2 inches platys = 4 × 2 = 8 inches

Seven 1 inches guppies = 7 × 1 = 7 inches

A 3 inches leach = 3 inches

The smallest capacity tank you can buy will be

V = 8 × 7 × 3

V = 168 cubic inches

Therefore, the smallest capacity of tank you can buy is 168 cubic inches

5 0

3 years ago

1+cos2A/cos2A = tan2A/tanA prove LHS=RHS

son4ous [18]

$RTP: \frac{1 + cos(2A)}{cos(2A)} = \frac{tan(2A)}{tan(A)}$

$LHS = \frac{1 + cos(2A)}{cos(2A)}$
$= \frac{1 + \frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}$
$= \frac{\frac{1 + tan^{2}(A) + 1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}$
$= \frac{2}{1 - tan^{2}(A)}$
$= \frac{2}{1 - tan^{2}(A)} \cdot \frac{tan(A)}{tan(A)}$
$= \frac{\frac{2tan(A)}{1 - tan^{2}(A)}}{tan(A)}$
$= \frac{tan(2A)}{tan(A)}$
$= RHS$ , as required.

3 0

4 years ago