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2 years ago

............Find .....X......

Mathematics

2 answers:

Stells [14]2 years ago

8 0

Looking from the picture it would be 50

LiRa [457]2 years ago

4 0

Answer:

x=50

Step-by-step explanation:

(x-10)+x+90=180

2x-10+90=180

2x+80=180

2x=100

x=50

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Find the area of the trapezoid below.

Pani-rosa [81]

Answer: 54 cm²

Step-by-step explanation: In this problem, we're asked to find the area of the trapezoid shown. A trapezoid is a quadrilateral with one pair of parallel sides.

The formula for the area of a trapezoid is shown below.

$Area =\frac{1}{2} (^{b} 1 + ^{b}2)h$

The b's represent the bases which are the parallel sides and h is the height.

So in the trapezoid shown, the bases are 6 cm and 12 cm and the height is 6 cm. Plugging this information into the formula, we have $\frac{1}{2} (6 cm +12cm)(6 cm)$ .

Next, the order of operations tell us that we must simplify inside the parentheses first. 6 cm + 12 cm is 18 cm and we have $\frac{1}{2}(18 cm)(6 cm)$ .

$\frac{1}{2} (18 cm)$ is 9 cm and we have 9 cm · 6 cm of 54 cm²

So the area of the trapezoid shown is 54 cm².

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3 years ago

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as and x,y point

Likurg_2 [28]

The equation of the parabola in the vertex form is y = (x - 3 $)^{2}$ - 5 with ( 3, -5) is the vertex of the parabola and 1 is the multiplier

In the above question, A parabolic equation is given as follows:

Y = x^2 - 6x + 4

The equation of the parabola in the vertex form is :

y = a (x - h $)^{2}$ + k

Where a is a multiplier in the equation and (h,k) are the coordinates of the vertex

So, in order to obtain this form, we will use the method of completing square :

Y = x^2 - 6x + 4

y = $x^{2}$ - 6x + (9 -9) + 4

y = (x - 3 $)^{2}$ + ( -9 + 4)

y = (x - 3 $)^{2}$ - 5

where, ( 3, -5) is the vertex of the parabola and 1 is the multiplier

Hence, The equation of the parabola in the vertex form is y = (x - 3 $)^{2}$ - 5 with ( 3, -5) is the vertex of the parabola and 1 is the multiplier

To learn more about, parabola, here

brainly.com/question/21685473

#SPJ1

8 0

1 year ago

in a test tube, there is 4 ML of oil and 6 ML of water. What volume percent of liquid in your test tube is oil? Show a ratio of

Nady [450]

3:2, easy.

the rest you dont need to read i just need 20 characters to "explain it well".

4 0

3 years ago

Which system of equation should you use to solve this problem

Alex17521 [72]

Last option is the correct answer.

Step-by-step explanation:

Given that

Let x stand for the solution a and y stand for the solution B

So,

According to statement that

30 gallons have to be made

$x+y=30$

Solution A contains 29 percent bromine => 0.29x

Solution B contains 53 percent =>0.53y

And the mixture of 30 gallons contains 45% bromine

$0.29x+0.53y=30*0.45\\0.29x+0.53y=13.5$

Following equations will be used for solving the problem

$x+y = 30\\0.29x+0.53y=13.5$

Hence,

Last option is the correct answer.

Keywords: Equations, Variables

Learn more about linear equations at:

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4 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