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3 years ago

Stanford form to y=Mx+b What is 6x-8y=32 in y=Mx+b form?

Mathematics

1 answer:

OverLord2011 [107]3 years ago

5 0

Answer:

y=3/4x-4 and ignore this stuff ejdndjsienensmsk

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What is the value of the 7 in 7 6 4

borishaifa [10]

The value of 7 is in the hundreds place

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3 years ago

Read 2 more answers

Find the equation of the linear function represented by the table below in slope intercept form.

grigory [225]

Answer:

y = -3x - 2

Step-by-step explanation:

slope (m) = (-8 - -5) / (2 - 1) ... difference of y/difference of x

m = -3

y = mx + b

y = -3x + b

for (3,-11) b = y + 3x = -11 + 3x3 = -2

equation: y = -3x - 2

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3 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago

(6+8)^2=6^2+8^2 statement true

ddd [48]

this is false

(6+8)^2= 196

6^2+8^2=100

They are not equal so this is a false statement

5 0

4 years ago

Insert grouping symbols 2 times 9+7

a_sh-v [17]

Answer:

2 times (9+7)

6 0

4 years ago