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3 years ago

A piece of cardboard is 24 centimeters long and 15 centimeters wide. What is the area?

Mathematics

2 answers:

galina1969 [7]3 years ago

8 0

$24*15=

360$

360 square centimeters

Jet001 [13]3 years ago

7 0

24 x 15 = 360 Length x width

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What is the value of x in the solution to the following system of equations?

Rainbow [258]

Solve and the solution would be (-1, 2)

x value = -1

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4 years ago

Someone, please help! I'm really confused

11111nata11111 [884]

The correct pair is option E, which is:

FH ≅ FH - reflexive property

ΔGFH ≅ ΔEFH - SAS theorem

<h3>What is the SAS Congruence Theorem?</h3>

The SAS theorems states that two triangles are congruent if they have two pairs of congruent sides and a pair of congruent included angles.

<h3>What is the Reflexive Property?</h3>

The reflexive property of geometry states that an angle or line will always be congruent to itself.

In the two column-proof, since FH = FH using the reflexive property, then both triangles are congruent to each other by the SAS congruence theorem.

The missing pair of reasons that completes the proof are:

FH ≅ FH - reflexive property

ΔGFH ≅ ΔEFH - SAS theorem

Learn more about the SAS theorem on:

brainly.com/question/2102943

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2 years ago

The equation of this circle is:

qaws [65]

Step-by-step explanation:

Equation : (x - 4)² + (y + 3)² = 4.

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3 years ago

Please dont ignore, Need help!!! Use the law of sines/cosines to find..

Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

$\sin{A} = \sin{103\textdegree{}}$ ,
The opposite side of angle A $a = BC = 26$ ,
The angle C is to be found, and
The length of the side opposite to angle C $c = AB = 6$ .

$\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}$ .

$\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}$ .

$\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}$ .

Note that the inverse sine function here $\sin^{-1}()$ is also known as arcsin.

By the law of cosine,

$c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C}$ ,

where

$a$ , $b$ , and $c$ are the lengths of sides of triangle ABC, and
$\cos{C}$ is the cosine of angle C.

For triangle ABC:

$b = 21$ ,
$c = 30$ ,
The length of $a$ (segment BC) is to be found, and
The cosine of angle A is $\cos{123\textdegree}$ .

Therefore, replace C in the equation with A, and the law of cosine will become:

$a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}$ .

$\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}$ .

For triangle ABC:

$a = 14$ ,
$b = 9$ ,
$c = 6$ , and
Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

$b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}$ .

$\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}$ .

$\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree$ .