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2 years ago

Television sizes are labeled by the measure of the diagonal. What size is the TV below?

Mathematics

1 answer:

alukav5142 [94]2 years ago

6 0

Answer:

55 inches.

Step-by-step explanation:

So, in the question above we are given the following data or parameters or information which will assist us in solving this kind of question efficiently. Here are the information needed below;

(1). "Television sizes are labeled by the measure of the diagonal.''

(2). "47.5 inches up top and the side is 27 inches" which are the length and the breadth respectively. Therefore, we will make use of the formula below In solving this question;

Size of the Television = √[(Length of the Television)^2 × (breadth of the television)^2].

Hence, the size of the Television = √( 47.5)^2 + (27)^2. = 54.63744137494.

Approximately:

The Size of the Television = 55 inches.

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Answer:

3=20

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Step-by-step explanation:

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3 years ago

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Answer:

It's irrational

Step-by-step explanation:

the square root of 27 is equal to:

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We know that $\sqrt{3}$ is an irrational number (but a real number), so $3\sqrt{3}$ is the same.

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3 years ago

Read 2 more answers

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Mandarinka [93]

Answer:

Step-by-step explanation:

4 x 5= 20

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2 years ago

Isaac keeps track of the amount of money in his savings certificate each month. He has accumulated the following data:

Dafna1 [17]

Answer:

d. The common ratio is 1.1

Step-by-step explanation:

To see if the data has a common ratio or common difference, we have to see if the division between them is equal(common ratio), or if the difference between them is equal(common difference).

In this case, since $60.5 - 55 \neq 55 - 50$ , it has a common ratio.

To find it, we divide consecutive terms. For example:

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So the correct answer is:

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2 years ago

Find the solution set of the following quadratic equations using the quadratic formula.

oksian1 [2.3K]

Answer:

<h3> 1) $x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$ </h3><h3> 2) $x_1=-\dfrac13\,,\quad x_2=-3$ </h3>

Step-by-step explanation:

$x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$

<h3>2)</h3><h3> $3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3$ </h3>

7 0

3 years ago