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3 years ago

Please help 10 points

Mathematics

1 answer:

Margarita [4]3 years ago

6 0

Answer:

The value of x is -2 and 2.

Step-by-step explanation:

When solve an expression, you have to make the expression equals to 0 by moving all the terms to one side :

x² - 3x - 1 = -3x + 3

x² - 3x - 1 + 3x - 3 = 0

x² - 4 = 0

Next, you have to factorize and solve it :

x² - 4 = 0

x² + 2x - 2x - 4 = 0

x(x + 2) - 2(x + 2) = 0

(x + 2)(x - 2) = 0

x + 2 = 0

x = -2

x - 2 = 0

x = 2

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Use elimination and explain

Temka [501]

Answer:

The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation. So if you have a system: x – 6 = −6 and x + y = 8, you can add x + y to the left side of the first equation and add 8 to the right side of the equation

4 0

3 years ago

Please help please please

monitta

Answer:

x=10

y=-2

(10,-2)

Step-by-step explanation:

If X=10, we can put 10 in for x in the equation.

30 + 5y = 20

Next, we need to solve to find y.

We will get y on its own so we subtract 30 from 20 to get

5y=-10

We then know that 5 goes into -10 -2 times.

y=-2

5 0

3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

If two angles are complementary then both angles must be acute true or false

Oxana [17]

Answer:

If two angles are acute angles, then they are complementary. Incorrect. The converse is “If two angles are complementary, then they are acute angles” which is true and the original statement was false because two acute angles are not always complementary.

Step-by-step explanation:

7 0

3 years ago

You can afford a $1350 per month mortgage payment. You’ve found a 30 year loan at 7% interest.

Mice21 [21]

A) 45 because 1350 divide by 30 equals 45. So 45 dollars u can afford

B) 1350 because 45 x 30 = 1350

C) 7% because it’s 7% to begin with

4 0

3 years ago