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vagabundo [1.1K]

3 years ago

15

HELPPPPPPPPPPPPIIDISKDKDJDNFN

2 answers:

Kisachek [45]3 years ago

6 0

Both are equal to each other

stepladder [879]3 years ago

5 0

$\displaystyle \bf\\\frac{1}{4} \cdot \frac{3}{4} = \frac{3}{4} \cdot \frac{1}{4}~~~\text{Equal because the multiplication is commutative.}\\\\ \\\frac{3}{4}\div\frac{2}{5} \neq \frac{2}{5}\div\frac{3}{4}~~~\text{Not equal because division is not commutative.}$

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omeli [17]

Answer:

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Step-by-step explanation:

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Andreyy89

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The weight of water is 62 1/2 lb per cubic foot. Water that weighs 350 lb will fill how many cubic feet?

valentinak56 [21]

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4 0

2 years ago

In the figure, a square is inside another bigger square.

evablogger [386]

Answer:

Part 1) The length of the diagonal of the outside square is 9.9 units

Part 2) The length of the diagonal of the inside square is 7.1 units

Step-by-step explanation:

step 1

Find the length of the outside square

Let

x -----> the length of the outside square

c ----> the length of the inside square

we know that

$x=a+b=4+3=7\ units$

step 2

Find the length of the inside square

Applying the Pythagoras Theorem

$c^{2}= a^{2}+b^{2}$

substitute

$c^{2}= 4^{2}+3^{2}$

$c^{2}=25$

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step 3

Find the length of the diagonal of the outside square

To find the diagonal Apply the Pythagoras Theorem

Let

D -----> the length of the diagonal of the outside square

$D^{2}= x^{2}+x^{2}$

$D^{2}= 7^{2}+7^{2}$

$D^{2}=98$

$D=9.9\ units$

step 4

Find the length of the diagonal of the inside square

To find the diagonal Apply the Pythagoras Theorem

Let

d -----> the length of the diagonal of the inside square

$d^{2}= c^{2}+c^{2}$

$d^{2}= 5^{2}+5^{2}$

$d^{2}=50$

$d=7.1\ units$

7 0

3 years ago

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Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the

Paha777 [63]

Answer:

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Step-by-step explanation:

Given

Tonya

$f(x) = 490 * 0.88^x$

Leo

$x \to g(x)$

$0 \to 480$

$2 \to 360$

$4 \to 470$

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

$f(x) = 490 * 0.88^x$

$f(0) = 490 * 0.88^0$

$f(0) = 490 * 1$

$f(0) = 490$

From Leo's table

$g(0) = 480$

By comparison;

$f(0) > g(0)$

i.e.

$490 > 480$

<em>So: Tonya's had the greater initial trade-in value</em>

Solving (b): The phone with lesser rate

An exponential function is:

$y = ab^x$

Where:

$b \to rate$

For Tonya

$b = 0.88$

For Leo, we have:

$(x_1,y_1) = (0,480)$

$(x_2,y_2) = (2,360)$

So, the equation becomes:

$y = ab^x$

$480 = ab^0$ and $360 = ab^2$

Solving $480 = ab^0$ , we have:

$480 = a * 1$

$480 = a$

$a= 480$

$360 = ab^2$ becomes

$360 = 480 * b^2$

Divide both sides by 480

$0.75 = b^2$

Take square roots

$0.87 = b$

$b=0.87$ -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88

6 0

3 years ago