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4 years ago

college algebra: Write an equation for a line g(t) perpendicular to h(t) =-3t+6 and passing through the point (-6,-1)

Mathematics

1 answer:

vladimir1956 [14]4 years ago

7 0

The answer should be g(t)=(1/3)t+1

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Which of the following statements justifies why the triangle shown below is

LenKa [72]

Answer:

A) 6^2 + 11^2 is not equal to 15^2

Step-by-step explanation:

To find out if a right triangle is a right triangle you have to use the Pythagorean Theorem, like it is used in A.

6^2 + 11^2 = 15^2

36 + 121 = 225

157 does not equal 225

So this is not a right triangle

4 0

3 years ago

Read 2 more answers

5/7=20/r<br><br> R=28<br><br> Can anyone explain how you get 28

Lesechka [4]

Answer:

Step-by-step explanation:

we are given that these two fractions are equal. however, 20/r is just given in numerators and denominators that are four times as large, even though it is still equal to 5/7. we can tell this because 20 is 4 times larger than 5, which means that R must be 4 times larger than 7, right? YES. so we take 7*4 to get 28.

8 0

3 years ago

Rodney created a graph of each given linear equation below, but Rodney made mistakes on each of his graphs. Explain Rodney’s err

iogann1982 [59]

Rodney's errors on each of the graphs given can be explained as in the explanation below.

<h3>Linear graphs of slope-intercept form equations</h3>

Rodney's error on each of the given linear graphs are as follows;

For y= 30x +10;

The slope of the graph is 30 and it's intercept is; 10. Hence, the graph should intersect/touch the y-axis at (0,10)

For y= (-1/2)x +5;

The slope of the given function is; -1/2 and hence, the graph should be a decreasing linear graph.

For y= x -2;

The slope and intercept of the given function are 1 and -2 respectively. On this note, the graphed line should intersect the y-axis at point (0, -2) and the graph should be an increasing linear graph.

Read more on linear graphs;

brainly.com/question/14323743

4 0

3 years ago

SOMEBODY HELP ME PLEASE

Karolina [17]

The exponential model indicates that hydropower energy generation has increased by 2.3% (0.023).

<h3>What is an exponential function?</h3>

An exponential function can be defined as a type of mathematical equation whose numerals are generated by a constant that's raised to the power of an argument.

Based on an exponential model for this data, we can deduce the following:

Mathematically, the exponential function for the given data is given by $y = 1276.6e^{0.023x}$

The exponential model indicates that hydropower energy generation has increased by 2.3% (0.023).

Hydropower energy generation in 25 years after 1970 was closest to 2270 TWh.

Because the correlation coefficient indicates a strong correlation between the model and the data, this prediction can be accepted with confidence interval of 98.7%.

Read more on exponential functions here: brainly.com/question/12940982

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5 0

2 years ago

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