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3 years ago

P(7, - 2); y= 7x+4= 4 PLEASE HELP

Mathematics

1 answer:

Mars2501 [29]3 years ago

3 0

Answer:

x= - 16

Step-by-step explanation:

substitute y=0 then multuply both sides move the variable to the left and change the sign.

You might be interested in

What property is this (8x5) x 10=8x(5x10)

sertanlavr [38]

Answer:

Associative

you can change (8*5) *10 to (8*10) *5

You can change 8*( 5*10) to 10* (8*5)

8 0

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

I’ve never been so confused, this is due in a few minutes, pls help if you can!

algol13

Given:

m(ar QT) = 220

m∠P = 54

To find:

The measure of arc RS.

Solution:

PQ and PT are secants intersect outside a circle.

If two secants intersects outside a circle, then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

$$\Rightarrow \angle P = \frac{1}{2} (m \ ar (QT) - m \ ar (RS ))$