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3 years ago

Find x. Assume that segments that appear tangent are tangent.

Mathematics

1 answer:

kipiarov [429]3 years ago

6 0

Assuming that CB is tangent, this is just a right triangle with a hypotenuse of 20 and a side of 12 so by the Pythagorean Theorem:

20^2=12^2+x^2

400=144+x^2

x^2=400-144

x^2=256

x=16

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Two cars travel along a highway at the same speed.

Mazyrski [523]

Answer:

30 mins

Step-by-step explanation:

<em>speed = distance/ time </em>

speed of car A

= 20/ 12 km/ min

= 5/ 3 km/ min

also, speed of car A = speed of car B

speed of car B = 5/ 3 km/ min = 50km / time taken (in mins)

$\frac{5}{3} = \frac{50}{t}$

$5t = 150$

$t = 30$

=> t = 30 mins

6 0

3 years ago

Triangles abc, dbe, and fbg are all symmetric about the y-axis. what are the coordinates of the centroid?

Drupady [299]

Answer:

The coordinates of the centroid are (0, 5)

Step-by-step explanation:

The coordinates of the centroid of a triangle whose vertices (x1, y1), (x2, y2), and (x3, y3) are ( $\frac{x1+x2+x3}{3}$ , $\frac{y1+y2+y3}{3}$ )

In Δ ABC

∵ A = (-20, 0), B = (0, 15), C = (20, 0)

∴ x1 = -20, x2 = 0, x3 = 20

∴ y1 = 0, y2 = 15, y3 = 0

∴ The centroid = ( $\frac{-20+0+20}{3}$ , $\frac{0+15+0}{3}$ ) = ( $\frac{0}{3}$ , $\frac{15}{3}$ ) = (0, 5)

∴ The coordinates of the centroid of Δ ABC are (0, 5)

In Δ DBE

∵ D = (-15, 0), B = (0, 15), E = (15, 0)

∴ x1 = -15, x2 = 0, x3 = 15

∴ y1 = 0, y2 = 15, y3 = 0

∴ The centroid = ( $\frac{-15+0+15}{3}$ , $\frac{0+15+0}{3}$ ) = ( $\frac{0}{3}$ , $\frac{15}{3}$ ) = (0, 5)

∴ The coordinates of the centroid of Δ DBE are (0, 5)

In Δ FBG

∵ F = (-5, 0), B = (0, 15), G = (5, 0)

∴ x1 = -5, x2 = 0, x3 = 5

∴ y1 = 0, y2 = 15, y3 = 0

∴ The centroid = ( $\frac{-5+0+5}{3}$ , $\frac{0+15+0}{3}$ ) = ( $\frac{0}{3}$ , $\frac{15}{3}$ ) = (0, 5)

∴ The coordinates of the centroid of Δ FBG are (0, 5)

∵ The three triangles are symmetric about the y-axis

→ That means they have the same centroid and it lies on the y-axis

∴ The coordinates of the centroid are (0, 5)

8 0

3 years ago

Complete the square and write in standard form. Show all work.What would be the conic section:CircleEllipseHyperbolaParabola

mote1985 [20]

ANSWER

This is an ellipse. The equation is:

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

EXPLANATION

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

$(a+b)^2=a^2+2ab+b^2$

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

$16x^2-32x=16(x^2-2x)$

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

$(x-1)^2=x^2-2x+1$

To maintain the equation, we have to subtract 1,

$16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16$

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

$16(x-1)^2-16+9y^2+72y+16=0$

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

$9y^2+72y=9(y^2+8y)$

Following the same reasoning as before, we have that the perfect binomial squared is,

$(y+4)^2=y^2+8y+16$

Remember to subtract the independent term to maintain the equation,

$9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144$

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

$16(x-1)^2-16+9(y+4)^2-144+16=0$

Add like terms,

$\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}$

Add 144 to both sides,

$\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}$

As we can see, this is the equation of an ellipse. Its standard form is,

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

$\begin{gathered} \frac{16(x-1)^2}{144}+\frac{9(y+4)^2}{144}=\frac{144}{144} \\ \\ \frac{(x-1)^2}{\frac{144}{16}}+\frac{(y+4)^2}{\frac{144}{9}}=1 \end{gathered}$

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

$\frac{(x-1)^2}{(\frac{12}{4})^2}+\frac{(y+4)^2}{(\frac{12}{3})^2}=1$

Note that we can simplify a and b,

$\frac{12}{4}=3\text{ and }\frac{12}{3}=4$

Hence, the equation of the ellipse is,

$\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1$

3 0

1 year ago

Terday. Priya successfully made 6 free throws. Today, she made 75% as

tatuchka [14]

Answer:

Step-by-step explanation:

Yesterday, Priya successfully made 6 free throws.

Today, she made 75% as many as yesterday.

We are asked to calculate the number of successful free throws that Priya made today.

So, the number will be $6 (1 + \frac{75}{100} ) = 10.5$ .

Since it is the number of successful free throws, so it will be 10 throws. ( Answer )

4 0

4 years ago

What is the x-coordinate of the solution to the following system of equations?

Alja [10]

Answer:

-1,2

Step-by-step explanation:

6 0

3 years ago