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Artyom0805 [142]
3 years ago
6

How do you find the exact value of sec θ if sin θ = -15/17 and 180 < θ < 270?

Mathematics
1 answer:
AnnyKZ [126]3 years ago
7 0

For 180^\circ, we expect to have \cos\theta. Then if \sin\theta=-\dfrac{15}{17}, we have

cos^2\theta+\sin^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac8{17}

\implies\sec\theta=\boxed{-\dfrac{17}8}

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The GCF of 10 and 6 is 2. May this will help you.

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After going through all of your expenses for the month, you can up with the expression -2(x+3) – 5(2x – 7). Simplify the express
svlad2 [7]

Answer:

-12x+29

Step-by-step explanation:

= -2(x+3) – 5(2x – 7)

= -2(x)+(-2)(3)-5(2x)-5(-7)

= -2x-6-10x+35

= -2x-10x-6+35

= -12x+29

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2 years ago
Y=1/2×-6 ×=-4 what is the solution to the system of equations
hodyreva [135]

Answer:

(-4, -8)

Step-by-step explanation:

Use the substitution method.  x = -4, so y = (1/2)x - 6 becomes:

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8 0
3 years ago
I’m stuck can some one help me
abruzzese [7]

Answer:

3) Midpoint is (-4,0.5)

Option A is correct.

4) Midpoint is (2.5,0)

Option B is correct.

5) The factors are (x+4)(x-7)

Option C is correct.

6) The factors are (x+4)(x+2)

Option A is correct.

Step-by-step explanation:

Question 3

Find midpoint of the following:

(2,-7), (-10,8)

The formula used to find midpoint is: Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )

We have x_1=2, y_1=-7, x_2=-10,y_2=8

Putting values and finding midpoint

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2-10}{2},\frac{-7+8}{2} )\\Midpoint=(\frac{-8}{2},\frac{1}{2} )\\Midpoint=(-4,0.5 )

So, Midpoint is (-4,0.5)

Option A is correct.

Question 4

Find midpoint of the following:

(2,-10), (3,10)

The formula used to find midpoint is: Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )

We have x_1=2, y_1=-10, x_2=3,y_2=10

Putting values and finding midpoint

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2+3}{2},\frac{-10+10}{2} )\\Midpoint=(\frac{5}{2},\frac{0}{2} )\\Midpoint=(2.5,0 )

So, Midpoint is (2.5,0)

Option B is correct.

Question 5

Factor each completely

x^2-3x-28

We will break the middle term and find factors

x^2-3x-28\\=x^2-7x+4x-28\\Taking\:common\\=x(x-7)+4(x-7)\\=(x+4)(x-7)

So, the factors are (x+4)(x-7)

Option C is correct.

Question 6

Factor each completely

x^2+6x+8

We will break the middle term and find factors

x^2+6x+8\\=x^2+4x+2x+8\\=x(x+4)+2(x+4)\\=(x+4)(x+2)

So, the factors are (x+4)(x+2)

Option A is correct.

7 0
2 years ago
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