All categories

2 years ago

Find the angle measure that of each missing angle

Mathematics

1 answer:

11111nata11111 [884]2 years ago

4 0

Step-by-step explanation:

122+v=180

v=180-122

v=58

w+61=180

w=180-61

w=119

61+58+y=180

119+y=180

y=180-119

y=61

61+x=180

x=180-61

x=119

j=180

You might be interested in

HELP!!! Andre reads 2 1/2 pages in 1/4 hour. What is the complex fraction and unit rate?

Gre4nikov [31]

Answer:

Step-by-step explanation:

2 1/2 = 5/2

5/2 x 4/1 = 20/1 = 20

(We are dividing fraction)

(keep change flip)

6 0

2 years ago

Solve the following system of equations by elimination. What is the value of x? x = 1.5 x = -1.5 x = 8 x = -8

Andru [333]

Using the elimination method, the value of x in the system of equations is calculated as: 8.

<h3>How to Solve a System of Equations by Elimination?</h3>

To solve a system of equations given using the elimination method, do the following:

Multiply 2x - 5y = 1 by 2 and multiply -3x + 2y = -18 by 5 to get the following:

4x - 10y = 2 --> eqn. 1

-15x + 10y = -90 --> eqn. 2

Add

-11x = -88

Divide both sides by -11

x = 8

Learn more about system of equations on:

brainly.com/question/14323743

#SPJ1

8 0

1 year ago

Use the description and table to graph the function, and determine the domain and range of f(x). Represent the domain and range

zmey [24]

The domain of the function is x > 0 and the range of the function is -∞ < f(x) < ∞

<h3>How to determine the domain?</h3>

The table represents a logarithmic function.

The domain represents the set of x values.

From the table, we can see that all the x values are positive values i.e. x > 0

Hence, the domain of the function is x > 0

<h3>How to determine the range?</h3>

From the table, the table outputs all zero, positive and negative numbers

This means that the range is the set of all real numbers.

Hence, the range of the function is -∞ < f(x) < ∞

Read more about domain and range at:

brainly.com/question/11787743

#SPJ1

7 0

2 years ago

Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

8 0

2 years ago

(1+tanx)/(sinx+cosx)=secx

jek_recluse [69]

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ is proved

<h3><u>Solution:</u></h3>

Given that,

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ ------- (1)

First we will simplify the LHS and then compare it with RHS

$\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}$ ------ (2)

$\text {We know that } \tan x=\frac{\sin x}{\cos x}$

Substitute this in eqn (2)

$=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

On simplification we get,

$=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}$

$=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}$

Cancelling the common terms (sinx + cosx)

$=\frac{1}{c o s x}$

We know secant is inverse of cosine

$=\sec x=R . H . S$

Thus L.H.S = R.H.S

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$

Hence proved

5 0

3 years ago