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Murljashka [212]
3 years ago
14

Two angles are complementary. They have measures of (7x + 2)° and (3x – 2)°, respectively. What is the value of x?

Mathematics
2 answers:
Nady [450]3 years ago
4 0
7x+2+3x-2=90\\
10x=90\\
x=9
marusya05 [52]3 years ago
3 0

Answer:

Option (c) is correct.

The value of x is 9.

Step-by-step explanation:

Given :  Two angles are complementary. They have measures of (7x + 2)° and (3x – 2)°, respectively.

We have to find the value of x.

Two angles are said to be complementary if the sum of measure of angle is 90°.

Given (7x + 2)° and (3x – 2)° are complementary angles.

That is sum of (7x + 2)° and (3x – 2)° is 90°

⇒ (7x + 2)° + (3x – 2)° = 90°

⇒ 7x + 2 + 3x - 2 = 90

⇒ 10x = 90

⇒ x = 9

Thus, the value of x is 9.

Hence, option (c) is correct.

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6 0
3 years ago
A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2
alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0
3 years ago
Apply Newton’s method to estimate the solution of x3 − x − 1 = 0 by taking x1 = 1 and finding the least n such that xn and xn +
ss7ja [257]

Solution :

Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

4 0
3 years ago
How does place value help you add decimal
gladu [14]
You line up the decimal points so that similar place values are lined up. it will be easier to add them.
3 0
3 years ago
Read 2 more answers
X squared minus 5x equals 0
Nadusha1986 [10]
Hope it helpssss…..best of luckkkkk

4 0
2 years ago
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