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3 years ago

Fill in the blank:

1 answer:

4 0

I can be wrong, but I think that the answer is: To find the values of p, q, r and s, you should start by finding all factor pairs of the leading coefficiant and constant term.

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What answer B? I’m confused

marishachu [46]

Answer:

about 35 mins

Step-by-step explanation:

5 0

3 years ago

HELP PLEASE I'm SOO CONFUSED

elena-s [515]

The point of intersection of these two lines is (2, 1)

The equation of a horizontal line that passes through the point (1, 1) is:

y = 1

The equation of a vertical line that passes through the point (2, 2) is:

x = 2

The point of intersection would be at:

x = 2, y = 1

The point of intersection of these two lines is (2, 1)

Find out more at: brainly.com/question/16530416

5 0

3 years ago

Solve the system by substitution. 4y = 5x -10y =-50

Trava [24]

Answer:

$x = -20 ; y =-5$

Step-by-step explanation:

Eqn. 1 ----> 4y = x

Eqn. 2 ----> 5x-10y = -50

(Simplifying eqn.2 further)

$=>5(x-2y)=-50$

$=>x-2y=\frac{-50}{5} =-10$

(Substituting the value of x from eqn. 1)

$=>4y-2y=-10$

$=>2y=-10$

$=>y =\frac{-10}{2} =-5$

Now, substituting the value of y in eqn. 1 ,

$x = 4*(-5)=-20$

7 0

3 years ago

PLEASE HELP ASAP!! WILL GIVE BRAINLEIST!!

GarryVolchara [31]

Answer:

The answer is B lol

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Alla [95]

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

5 0

3 years ago