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3 years ago

AB and BC are tangents. Find x.

Mathematics

1 answer:

Whitepunk [10]3 years ago

3 0

Answer:

x=74

Step-by-step explanation:

If two tangent segments are drawn from the same external point, then the segments are equal.

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I HAVE ASKED THIS THREE TIMES. PLEASE ANSWER <3

Andru [333]

Answer:

Step-by-step explanation:

For Question 3, we are simply taking an input for the function, as a value of x and solving the equation. For part a, we substitute 3/14 into the first function, and solve it:

f(x) = 7(3/14) + 2

f(x) = 21/14 + 2

f(x) = 49/14

f(x) = 7/2

For part b, we take the input of -3 into the second function and solve the equation:

h(x) = 4(-3)^2

h(x) = 4(9)

h(x) = 36

For Question 4, we are simply solving this equation by isolating the x variable. First, we simplify the equation to 4-5x+15+2x = -2 and simplify this again to -3x+19 = -2. Now, we can subtract 19 from both sides of the equation to get -3x = -21. Lastly, we isolate the x variable by dividing both sides of this equation by -3, to get x = 7.

8 0

3 years ago

Read 2 more answers

K=6x+10, solve for k

kicyunya [14]

Answer:

x=(10-k)÷6

Step-by-step explanation:

6x=10-k

x=(10-k)÷6

4 0

3 years ago

In this circle, the area of sector COD is 50.24 square units. The radius of the circle is units , and units.

maks197457 [2]

Let angle be alpha

$\\ \rm\Rrightarrow Area=50.24$

$\\ \rm\Rrightarrow \dfrac{\alpha}{360}\pi r^2=50.24$

$\\ \rm\Rrightarrow \alpha 3.14r^2=18086.4$

$\\ \rm\Rrightarrow \alpha r^2=5760$

$\\ \rm\Rrightarrow r^2=\dfrac{5760}{\alpha}$

$\\ \rm\Rrightarrow r=\sqrt{\dfrac{5760}{\alpha}}$

If you have angle measure in degree put in place of Alpha and get the value elaw this is our final answer

7 0

2 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

Subtract 7 from the product of 4 and a number results in the number added to 29 is what

german

4x-7=29 < do you understand how to set that up? Now, the goal is to “get x by itself.”

4x-7=29< now add the 7 to the 29. When you cross an equals sign you change the (+/-) sign of the number you’re moving.

4x= 36< now you will just divide by 4 on both sides. This will give you x by itself on the left side of the equals sign.

x=36/4 which is 9

X=9 :))

7 0

3 years ago

Read 2 more answers