All categories

3 years ago

Name the segment that is parallel to the given segment

Mathematics

1 answer:

ioda3 years ago

5 0

Answer:

Step-by-step explanation:

Parallel lines never touch. Line yz would never touch line fg.

You might be interested in

What is the approximate solution to the equation 2^t =38

timurjin [86]

2^t =38

take the log of each side

log ( 2^t) = log (38)

the exponent gets multiplies

t log (2) = log (38)

divide by log 2

t = log(38)/log (2)

t≈5.2479

3 0

3 years ago

Find the measures of two complementary angles, ÐA and ÐB, if mÐA = (7x + 4)° and mÐB = (4x + 9)°. Measure of angle A = ... Measu

Westkost [7]

Answer:

The measure of angle A is 53 degrees

The measure of angle B is 37 degrees

Step-by-step explanation:

Here in this question, we are interested in calculating the measure of the two angle.

When how angles are complementary, it means the sum of both equals 90 degrees

mathematically ;

(7x + 4) + (4x + 9) = 90

7x + 4x + 4 + 9 = 90

11x + 13 = 90

11x = 90-13

11x = 77

x = 77/11 = 7

The measure of angle A = 7x + 4 = 7(7) + 4 = 49 + 4 = 53 degrees

The measure of B = 4x + 9 = 4(7) + 9 = 28 + 9 = 37 degrees

3 0

3 years ago

An office supply company is shipping a case of wooden pencils to a store. There are 64 boxes of pencils in a case. If each box o

I am Lyosha [343]

The wooden pencils weigh 10 pounds all together.

3 0

3 years ago

Read 2 more answers

Given that cos theta = -12/13 and theta is in the second quadrant, find this function. sin theta=

elena-14-01-66 [18.8K]

5/13

The x value is negative (second quadrant). 13 is the hypotenuse. So the missing side is 13^2 = (-12)^2 + y^2.

Sin theta is the opposite (5) over hypotenuse (13).

3 0

3 years ago

Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt

tino4ka555 [31]

Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

$f(x)= \sqrt{\dfrac{x+1}{x-1}}$

$g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}$

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

$\dfrac{x+1}{x-1}>0$

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

$x+1\leq 0\Rightarrow x\leq -1$

$x-1\leq 0\Rightarrow x\leq 1$

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

4 0

4 years ago