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1 year ago

Solve for x A. 4 B. 7 C. 1 D. 9

Mathematics

1 answer:

neonofarm [45]1 year ago

4 0

The value of x when the secants intersect is 4

<h3>How to find side when two secant intersect?</h3>

A secant of a circle is a line that connects two distinct points on a curve.

The secant touches two sides of the circumference of a circle.

Therefore, using secant rule,

4(x + 2 + 4) = 5(x - 1 + 5)

Hence,

4(x + 6) = 5(x + 4)

Hence, open the brackets

4x + 24 = 5x + 20

Therefore, subtract 4x from both sides

4x - 4x + 24 = 5x - 4x + 20

24 = x + 20

subtract 20 from both sides

24 - 20 = x + 20 - 20

x = 4

Therefore, the value of x is 4.

learn more on secant here: brainly.com/question/19906313

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Solve the equation. \dfrac5{13}=t-\dfrac{6}{13} 13 5 =t− 13 6 start fraction, 5, divided by, 13, end fraction, equals, t, mi

OleMash [197]

Answer:

11 /13 = t

Step-by-step explanation:

5/13 = t -6/13

Add 6/13 to each side

5/13 + 6/13 = t -6/13+ 6/13

11 /13 = t

4 0

2 years ago

Read 2 more answers

Generate the next three terms of each geometric sequence

diamong [38]

Answer:

For the first one: -2, -1, 0

For the second one: 10, 15, 20

7 0

2 years ago

The area of a rectangle is 72 cm2. State the dimensions of the rectangle with the least perimeter. Calculate the perimeter.

Nitella [24]

We know that
<span>Of all the rectangles having same area, square has the least perimeter
</span>
area of square=x²
where
x is the length side of the square
A=72 cm²
72=x²
x=√72-----> x=6√2 cm

Perimeter of square=4*x------> 4*(6√2)-----> P=24√2 cm
P=33.94 cm

the answer is
the perimeter is equal to 24√2 cm or 33.94 cm

5 0

2 years ago

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

2 years ago

In constructing a box-and-whisker plot for the data set below, at which values would you draw the left and right sides of the bo

finlep [7]

Lets put these numbers in order
1,2,3,3,4,5,6,7,8,8,9

Q1 = 3
Q2 (median) = 5
Q3 = 8

the box starts at Q1 and ends at Q3....so left side is 3 and right side is 8
u r correct...it is A

7 0

3 years ago