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labwork [276]
3 years ago
13

What is the x-intercepts of the graph of y = 12x-5x-2

Mathematics
2 answers:
hammer [34]3 years ago
8 0
Y = 12x - 5x - 2

first simplify the equation by subtracting like terms (in this case):

12x - 5x = 7x

y = 7x - 2

Since you are finding the x, you must isolate the x. Do the opposite of PEMDAS.(Note: because there is a equal sign, what you do to one side, you do to the other)

y = 7x - 2

y (+2) = 7x - 2 (+2)
y + 2 = 7x
(y + 2)/7 = 7x/7

x = (y + 2)/7


x = (y + 2)/7 is your answer

hope this helps
PilotLPTM [1.2K]3 years ago
6 0
X intercept = when the graph cuts the x-axis = when y = 0


y = 12x^2 -5x -2

when y = 0
12x^2 -5x -2 = 0
(3x - 2) (4x + 1) = 0
x = 2/3 or x = 1/4

So the x-intercept are (2/3, 0) and (1/4, 0)

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Jen scored 80 on a test. 16 of the questions she answered were correct. How many questions did the test have?
Dmitry [639]
16 questions right = 80%
So that means Jen missed 4 questions.
Hope this helped.
8 0
3 years ago
Find the first term and common difference of an AP, whose 7th term is 29/6 and 15th term is 11/2.​
777dan777 [17]

Answer:

The first term is 13/3 and the common difference is d = 1/12

The formula is a(n) = 13/3 + (1/12)(n - 1)        

Step-by-step explanation:

The general equation for an arithmetic progression is:

a(n) = a(1) + d(n - 1), where d is the common difference/

Case 1:  n = 7:  29/6 = a(1) + d(7 - 1), or 29/6 = a(1) + d(6)

Case 2:  n = 15:  11/2 = a(1) + d(15 - 1) =  a(1) + d(14)

Then our system of linear equations is:

a(1) + 6d = 29/6

a(1) + 14d = 11/2

Let's solve this by elimination by addition and subtraction.  Subtract the first equation from the second.  We get:

                                                             

Substituting 1/12 for d in the first equation, we get:

a(1) + 14(1/12) = 11/2 or 66/12 (using the LCD 12)

Then a(1) = 66/12 - 14/12 = 52/12 = 13/3

The first term is 13/3 and the common difference is d = 1/12

The arithmetic sequence formula for this problem is thus:

a(n) = 13/3 + (1/12)(n - 1)                                                                          8

5 0
3 years ago
Whats the solution ​
Sergio039 [100]

Answer:

Step-by-step explanation:

To find the center and the radius, we need to put the equation in the (x-a)^2+(y-b)^2=r^2 [a is the x-coordinate, b is the y-coordinate, and r is the radius].

(x+6)^2+(y+7)^2 - 36 - 49 + 69 = 0

(x+6)^2+(y+7)^2 - 16 = 0

(x+6)^2+(y+7)^2 = 16

(x+6)^2+(y+7)^2 = 4^2

The center is (-6,-7) and the radius is 4.

Hope this helps!

5 0
2 years ago
given that sin theta= 1/4, 0 is less than theta but less than pi/2, what is the exact value of cos theta
lapo4ka [179]

Answer:

\cos{\theta} = \frac{\sqrt{15}}{4}

Step-by-step explanation:

For any angle \theta, we have that:

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

Quadrant:

0 \leq \theta \leq \frac{\pi}{2} means that \theta is in the first quadrant. This means that both the sine and the cosine have positive values.

Find the cosine:

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

(\frac{1}{4})^{2} + (\cos{\theta})^{2} = 1

\frac{1}{16} + (\cos{\theta})^{2} = 1

(\cos{\theta})^{2} = 1 - \frac{1}{16}

(\cos{\theta})^{2} = \frac{16-1}{16}

(\cos{\theta})^{2} = \frac{15}{16}

\cos{\theta} = \pm \sqrt{\frac{15}{16}}

Since the angle is in the first quadrant, the cosine is positive.

\cos{\theta} = \frac{\sqrt{15}}{4}

3 0
3 years ago
If (z°z)(x)=1/16x , what is z (x)?
vivado [14]
<span>assume z = ax for simplicity z(z) = a(ax) = a^2x let a^2x = 1/16x and solve for a </span>
6 0
3 years ago
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