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2 years ago

12

Which equation shows a proportional relationship?

2 answers:

xz_007 [3.2K]2 years ago

6 0

B is because any form of y=Mx +c is proportional

tatyana61 [14]2 years ago

6 0

I think the answer is b

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Find the missing side lengths. Leave your answers as radicals in simplest form. Please help!! Due today

Nataly_w [17]

Answer:

$m = 10\sqrt 3$

$n = 10$

Step-by-step explanation:

Required

Find m and n

Considering the given angle, we have:

$\sin(60) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60) = \frac{m}{20}$

Make m ths subject

$m = 20 * \sin(60)$

$\sin(60) =\frac{\sqrt 3}{2}$

So, we have:

$m = 20 *\frac{\sqrt 3}{2}$

$m = 10\sqrt 3$

Considering the given angle again, we have:

$\cos(60) = \frac{Adjacent}{Hypotenuse}$

This gives:

$\cos(60) = \frac{n}{20}$

Make n the subject

$n = 20 * \cos(60)$

$\sin(60) =\frac{1}{2}$

So, we have:

$n = 20 *\frac{1}{2}$

$n = 10$

5 0

2 years ago

PLZ HELP !! ABC OR D

kompoz [17]

Answer:

A

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In a cross-country run organized by a school, the students who completed the run

coldgirl [10]

280 students have a total of 560 points

There 60 more boys(x+60) complete than girls(x)

2*(x+60)+2*x= 560

number of girl =x =110

girl point =110*2= 220

6 0

2 years ago

)) Evaluate the expression for b = 7.<br> 11b - 75 =

Wewaii [24]

Answer:

The answer is 2.

Step-by-step explanation:

11(7) - 75

77-75

=2

7 0

3 years ago

How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

lisov135 [29]

Answer:

$300\; \rm lb$ .

Step-by-step explanation:

Let $x$ represent the mass (in pounds) of that $15\%$ copper alloy required, such that the final mixture would contain $25.5\%$ copper by mass.

Consider: if $x$ pounds of that $15\%$ copper alloy is mixed with $700$ pounds that $30\%$ copper alloy, what would be the mass of copper in the mixture?

Mass of copper in $x$ pounds of that $15\%$ copper alloy: $(0.15\, x)\; \rm lb$ .
Mass of copper in $700$ pounds of that $30\%$ copper alloy: $700 \times 0.30 = 210\; \rm lb$ .

Therefore, the mixture would contain $(210 + 0.15\, x) \; \rm lb$ of copper.

The mass of that mixture would be $(700 + x)\; \rm lb$ . The mass fraction of copper in that mixture would be:

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%$ .

This ratio is supposed to be equal to $25.5\%$ . These two pieces of equations combine to give an equation about $x$ :

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%$ .

$\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255$ .

Simplify and solve for $x$ :

$210 + 0.15\, x= 0.255\, (700 + x)$ .

$(0.255 - 0.15)\, x= 210 - 0.255 \times 700$ .

$\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300$ .

Therefore, $300\; \rm lb$ of that $15\%$ alloy would be required.

4 0

2 years ago