HCPS Algebra

T=3 and t=5 to determine if the expression 4(t+3) and 4 t+12 are equivalent

Mademuasel [1]

No they’re not equivalent because 4(t+12) =68 and 4t+12=32

7 0

3 years ago

If 3 cups of rice serves 5 people, how many cups do you need to serve 13 people

elena-14-01-66 [18.8K]

First let's find out how many cups serve one person.

3 ÷ 5 = 1.6

1.6 of a cup serves one person.

1.6 × 13 = 20.8

It takes 20.8 (21 rounded) to serve 13 people.

3 0

3 years ago

Write the equation in standard form<br><br> y = -10x + 6

Mumz [18]

Step (1) Flip the equation

−10x+6=y

Step (2) Add -6 to both sides

−10x+6+−6=y+−6

−10x=y−6

Step (3) Divide both sides by -10

−10x−10=y−6−10

ANSWER
$x = \frac{ - 1}{10} y + \frac{3}{5}$

5 0

3 years ago

Point A is located at (-2, 2), and point M is located at (1,0). If point M is the midpoint of AB, find the location of point B.

bezimeni [28]

Answer:

B. $B = (4,-2)$

Step-by-step explanation:

GIven that $A = (-2, 2)$ and $M = (1, 0)$ , and that point M is the midpoint of AB, the midpoint can be determined as a vectorial sum of A and B. That is:

$M = \frac{1}{2}\cdot A + \frac{1}{2}\cdot B$

The location of B is now determined after algebraic handling:

$\frac{1}{2}\cdot B = M - \frac{1}{2}\cdot A$

$B = 2\cdot M -A$

Then:

$B = 2\cdot (1,0)-(-2,2)$

$B = (2\cdot 1, 2\cdot 0)-(-2,2)$

$B = (2,0) -(-2,2)$

$B = (4,-2)$

Which corresponds to option B.

5 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

3 years ago