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3 years ago

A building is 75 feet tall. On a scale model , 5cm will represent 3 feet How many cm tall will the model be

Mathematics

2 answers:

Alex787 [66]3 years ago

8 0

Just do 75 divided by 3 and that gives you 25 do 25x5 and that gives you 125 the answer is 125.

Answer 125cm

zysi [14]3 years ago

7 0

125 cm idk if this is right

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vlada-n [284]

A)
145 - 5d
5d = 145, set equal

29 days to lay all the track

b)
5*19 = 95

145-95 = 50

After 19 days, they have 50 miles to go.

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3 years ago

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Evgesh-ka [11]

Answer:

-52, and the opposite of this is 52.

Step-by-step explanation:

If we are losing 52 pounds, then our number is -52 since we are losing 52 pounds (adding a negative number to a positive number is the same as subtracting that number).

The opposite of a number is when we negate the number, or multiply it by -1.

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3 years ago

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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.

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3 years ago

Line $\ell_1$ is horizontal and passes through $(1,1)$. Line $\ell_2$ is vertical and passes through $(2,2)$. Find the point of

borishaifa [10]

Answer:

(2,1)

Step by step explanation:

Please find the attachment.

We have been given that line $l_1$ is horizontal and passes through point (1,1). Line $l_2$ is vertical and passes through point (2,2). We are asked to find point of intersection of line $l_1$ and line $l_2$ .

Line $l_1$ is a horizontal line so y-coordinate will be 1 for any value of x. Line $l_2$ is a vertical line, so x-coordinate will be 2 for any value of y.

Therefore, intersection point of both lines will include x-coordinate of line $l_2$ and y-coordinate of line $l_1$ that is (2,1).

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3 years ago

Can someone help me with this question

Flura [38]

The answer is A or c I’m probs wrong

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3 years ago