All categories

3 years ago

What is the measure of x?

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

Answer:

x=4

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

(x+5) *x = 6^2

x^2 +5x = 36

Subtract 36 from each side

x^2 +5x - 36 = 0

Factor

( x-4) (x+9) = 0

Using the zero product property

x-4 = 0 x+9 =0

x = 4 x=-9

Cannot be negative since that is negative length

x=4

You might be interested in

Let f(x) = 4x2 + 6 and let g(x) = 8x – 6. Find f(g(x)) and g(f(x)).

Artyom0805 [142]

Answer:

f(g (x))= 32x-18

g (f (x))=32x2+42

Step-by-step explanation:

f (g (x))= 4 (8x-6)+6

32x-24-6

32x-18

g (f (x))= 8(4x2+6)-6

32x2+48-6

32x2+42

4 0

3 years ago

Find the missing length indicated. х 12 9 X = units

madam [21]

Answer:

Step-by-step explanation:

9/12=12/x

x=16

8 0

2 years ago

A grid shows the position of a subway stop at your house. The subway stop is located at (-5, 2) and your house is located (-9,9)

aleksley [76]

$A(-5;\ 2);\ B(-9;\ 9)\\\\|AB|=\sqrt{(-9-(-5))^2+(9-2)^2}=\sqrt{(-4)^2+7^2}=\sqrt{16+49}\\\\=\sqrt{65}\approx8\ (units)$

3 0

3 years ago

Evan saves 7 dollars each week. let d represents the total amount of money in dollars Evan saves after w weeks.

DerKrebs [107]

D=7
D to W = 19
Gives total savings on Week W $133

8 0

2 years ago

The perimeter of a triangular field is 120m. Two of the sides are 21m and 40m.Calculate the largest angle of the field.

Mama L [17]

Answer:

the largest angle of the field is 149⁰

Step-by-step explanation:

Given;

perimeter of the triangular filed, P = 120 m

length of two known sides, a and b = 21 m and 40 m respectively

The length of the third side is calculated as follows;

a + b + c = P

21 m + 40 m + c = 120 m

61 m + c = 120 m

c = 120 m - 61 m

c = 59 m

↓ ↓

A → → → → → → → → → → → C

Consider ABC as the triangular field;

Angle A is calculated by applying cosine rule;

$a^2 = b^2 + c^2 - 2bc \ Cos A\\\\Cos \ A = \frac{b^2 + c^2 - a^2}{2bc} \\\\Cos \ A = \frac{40^2 + 59^2 - 21^2}{2 \times 40 \times 59} \\\\Cos \ A = 0.983\\\\A = Cos ^{-1} (0.983)\\\\A = 10.6 \ ^0$

Angle B is calculated as follows;

$Cos \ B = \frac{a^2 + c^2 - b^2}{2ac} \\\\Cos \ B = \frac{21^2 + 59^2 - 40^2}{2 \times 21 \times 59} \\\\Cos \ B = 0.937\\\\B= Cos ^{-1} (0.937)\\\\B = 20.5 \ ^0$

Angle C is calculated as follows;

$Cos \ C = \frac{a^2 + b^2 - c^2}{2ab} \\\\Cos \ C = \frac{21^2 + 40^2 - 59^2}{2 \times 21 \times 40} \\\\Cos \ C = -0.857\\\\C = Cos ^{-1} (-0.857)\\\\C = 149\ ^0$

Therefore, the largest angle of the field is 149⁰.

8 0

3 years ago