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8_murik_8 [283]

2 years ago

8

The rectangle shown has a perimeter of 56 cm and the given area. Its length is 8 times more than three times its width. Write an

d solve a system of equations to find the dimensions of the rectangle.

1 answer:

Anna35 [415]2 years ago

3 0

Answer:

Dimensions of rectangle = <u>23 cm x 5 cm</u>

Step-by-step explanation:

<u>Equations we can form</u>

Perimeter of 56 cm

⇒ 2L + 2W = 56

⇒ L + W = 28

Length is 8 more than 3 times its width

⇒ L = 3W + 8

<u>Substitute in previous equation</u>

⇒ 3W + 8 + W = 28

⇒ 4W = 20

⇒ W = <u>5 cm</u>

<u></u>

⇒ L = 28 - 5 = <u>23 cm</u>

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20 has many factors, including 4&5.

4 is one less than 5, so 4 is the length and 5 is our width.

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Read 2 more answers

Determine whether each expression is equivalent to 49^2t – 0.5.

vampirchik [111]

Answer:

None of the expression are equivalent to $49^{(2t - 0.5)}$

Step-by-step explanation:

Given

$49^{(2t - 0.5)}$

Required

Find its equivalents

We start by expanding the given expression

$49^{(2t - 0.5)}$

Expand 49

$(7^2)^{(2t - 0.5)}$

$7^2^{(2t - 0.5)}$

Using laws of indices: $(a^m)^n = a^{mn}$

$7^{(2*2t - 2*0.5)}$

$7^{(4t - 1)}$

This implies that; each of the following options A,B and C must be equivalent to $49^{(2t - 0.5)}$ or alternatively, $7^{(4t - 1)}$

A. $\frac{7^{2t}}{49^{0.5}}$

Using law of indices which states;

$a^{mn} = (a^m)^n$

Applying this law to the numerator; we have

$\frac{(7^{2})^{t}}{49^{0.5}}$

Expand expression in bracket

$\frac{(7 * 7)^{t}}{49^{0.5}}$

$\frac{49^{t}}{49^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{49^{t}}{49^{0.5}}$ becomes

$49^{t-0.5}}$

This is not equivalent to $49^{(2t - 0.5)}$

B. $\frac{49^{2t}}{7^{0.5}}$

Expand numerator

$\frac{(7*7)^{2t}}{7^{0.5}}$

$\frac{(7^2)^{2t}}{7^{0.5}}$

Using law of indices which states;

$(a^m)^n = a^{mn}$

Applying this law to the numerator; we have

$\frac{7^{2*2t}}{7^{0.5}}$

$\frac{7^{4t}}{7^{0.5}}$

Also; Using law of indices which states;

$\frac{a^{m}}{a^n} = a^{m-n}$

$\frac{7^{4t}}{7^{0.5}}$ = $7^{4t - 0.5}$

This is also not equivalent to $49^{(2t - 0.5)}$

C. $7^{2t}\ *\ 49^{0.5}$

$7^{2t}\ *\ (7^2)^{0.5}$

$7^{2t}\ *\ 7^{2*0.5}$

$7^{2t}\ *\ 7^{1}$

Using law of indices which states;

$a^m*a^n = a^{m+n}$

$7^{2t+ 1}$

This is also not equivalent to $49^{(2t - 0.5)}$

6 0

3 years ago

Find x please help thank you

Angelina_Jolie [31]

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Step-by-step explanation:

To find x, we would have to find the hypotenuse of the 45-45-90 triangle. First, we would have to find the hypotenuse by using the 30-60-90 triangle on top to find it.

For a 30-60-90 triangle, the hypotenuse is 2x in length. the x is the same in all sides. All you would have to do is to plug it in. The leg opposite of 60° is x√3 in length. the leg opposite of 30° is x in length.

Since we know that 7 is opposite of the 30° angle, we know that x is 7. Across fron 60° is the hypotenuse of the 45-45-90 triangle. That leg is x√3. We plug in x=7 and get 7√3.

The hypotenuse of the 45-45-90 triangle is x√2 and the legs are both x. We can set 7√3 equal to x√2 to find x of the missing side.

7√3=x√2 [divide both sides by √2]

x=(7√6)/2

Now, we know x=(7√6)/2.

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago