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3 years ago

X : 18 : 30 =36 : 27 : y, find the value of x and y

Mathematics

1 answer:

svetlana [45]3 years ago

4 0

X = 24
y = 45

Mark brainliest please

Hope this helps

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When reading a word problem, there are clues that will help you to determine whether you need to find the surface area or the vo

Alex17521 [72]

The options are as follows

(1) An object that is filled

(2) Covering the outside of an object

(3) An object that is solid

(4) Cubic units

(5) Square units

Answer:

(1) An object that is filled

(3) An object that is solid

(4) Cubic units

Step-by-step explanation:

Clues that applies

(1) An object that is filled: By filling means that object has that amount of volume that is being filled completely.

(3) An object that is solid is likely a clue to finding the volume of an irregularly shaped solid.

(4) Cubic units, volume is measured in cubic units.

Clues that Doesn't Apply

(2) Covering the outside of an object is a clue to finding the surface area of an object.

(5) Square units is a measure of area not volume.

3 0

3 years ago

Read 2 more answers

An object falls from a position of rest and

kumpel [21]

Answer:

About 4.1323 meters.

Step-by-step explanation:

We can use the following kinematic equations:

$\displaystyle v_f = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2$

We want to determine the distance the object has dropped after falling from rest and reaching an instantenous speed of 9 m/s.

We are given that the acceleration due to gravity is 9.8 m/s².

Using this fact and the first equation, find the time for which it took the object to reach 9 m/s. Note that the initial velocity is 0 m/s since the object started from rest.

$\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s$^2$})t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}$

To find how far the object dropped, we can use the second equation:

$\displaystyle\begin{aligned} \Delta d & = v_0 t + \frac{1}{2} at^2 \\ \\ & = (0\text{ m/s})(0.9184 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s$^2$})(0.9184 \text{ s})^2 \\ \\ & = 4.1323 \text{ m} \end{aligned}$

In conclusion, the object would have dropped about 4.1323 meters.

6 0

3 years ago

I need help with pre calculus.

otez555 [7]

Question 1

<h3>Answer: Choice A) -7</h3>

-----------------

Explanation:

To compute f(g(2)), we'll first need to find g(2).

Locate 2 in the x row. Then read straight down until you get to the g(x) row. You should find that g(2) = 4

This means f(g(2)) becomes f(4). We then repeat the same process as before. Locate 4 in the x row, read straight down til you get to the f(x) row to find that f(4) = -9

Overall, we can say f(g(2)) = -9

We do the same set of steps to compute g(f(-1)). You should find that f(-1) = -1 and g(f(-1)) = g(-1) = -2

Lastly, we subtract the results we got:

f(g(2)) - g(f(-1)) = -9 - (-2) = -9+2 = -7

==============================================

Question 2

<h3>Answer: Choice B) 0</h3>

-----------------

Explanation:

If you were to graph each equation, you should get what you see below.

The red and blue curves do not intersect in any way. We need an intersection point to have a solution. Because there are no intersections, we don't have any solutions. The system is considered inconsistent.

==============================================

Question 3

<h3>Answer: Choice B</h3><h3> $-2x^2+250x-2000$ </h3>

-----------------

Explanation:

x = number of scooters made and sold

Each scooter costs $150 to make. If you made x of them, then it costs 150x dollars. Then tack on the fixed cost of $2000 to get the expression $150x+2000$

The cost function is $C(x) = 150x+2000$ . We'll subtract this from the revenue function R(x) to get the profit P(x)

$\text{Profit} = \text{Revenue} - \text{Cost}\\\\P(x) = R(x) - C(x)\\\\P(x) = (400x-2x^2) - (150x+2000)\\\\P(x) = 400x-2x^2-150x-2000\\\\P(x) = -2x^2+250x-2000\\\\$

==============================================

Question 4

<h3>Answer: Choice B</h3>

$\left\{x\in \mathbb{R} \big| \ x \ne -2, x \ne 1\right\}$

-----------------

Explanation:

$g(x) = \frac{1}{x+1}\\\\g(f(x)) = \frac{1}{f(x)+1}\\\\g(f(x)) = \frac{1}{x^2+x-3+1}\\\\g(f(x)) = \frac{1}{x^2+x-2}\\\\g(f(x)) = \frac{1}{(x+2)(x-1)}\\\\(g\circ f)(x) = \frac{1}{(x+2)(x-1)}\\\\$

I started with the outer function g(x) and replaced each x with f(x) in the second step. Then I plugged in f(x) = x^2+x-3 and simplified. Factoring the denominator is helpful to determine where the denominator becomes zero. Specifically, it happens when x = -2 and x = 1. These x values must be kicked out of the domain to avoid a division by zero error.

So we start off with the real number line, and poke holes at x = -2 and x = 1.

The notation $x \in \mathbb{R}$ means x is a real number. Then we tack on $x \ne -2, x \ne 1$

Overall, the domain is $\left\{x\in \mathbb{R} \big| \ x \ne -2, x \ne 1\right\}$

Let's see what happens if we tried plugging x = -1 into this composite function:

$(g \circ f)(x) = \frac{1}{(x+2)(x-1)}\\\\(g \circ f)(-1) = \frac{1}{(-1+2)(-1-1)}\\\\(g \circ f)(-1) = \frac{1}{(1)(-2)}\\\\(g \circ f)(-1) = \frac{1}{-2}\\\\(g \circ f)(-1) = -0.5\\\\$