All categories

3 years ago

PLEASE HELP ME. I NEED HELPPPPPPP ASAP

Mathematics

2 answers:

IceJOKER [234]3 years ago

8 0

Answer:

the answer is d. I hope this helps but uh slope is rise over run so you're looking for a line that goes up one and over two and passes through x4 and y2

mina [271]3 years ago

6 0

Answer:D

you have to calculate the slope

You might be interested in

Use the drawing tool(s) to form the correct answer on the provided grid.

solniwko [45]

She is starting at 4000, so that is the y-intercept of the graph. She is descending at 40 feet per minute, so her rate of change or slope is -40. The graph would look like this:

4 0

3 years ago

Read 2 more answers

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

J(x)=−45x+7; j(x)=−5

ASHA 777 [7]

Answer:

x = 4/15

Step-by-step explanation:

Step 1: Define

j(x) = -45x + 7

j(x) = -5

Step 2: Substitute and Evaluate

-5 = -45x + 7

-12 = -45x

x = 4/15

3 0

4 years ago

Estimate the measure of angle 1

fenix001 [56]

Answer:

most likely (c) 140 degrees

8 0

3 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS PLS

Elza [17]

Answer:

(-3,5)

Step-by-step explanation:

5 0

2 years ago