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xeze [42]
3 years ago
14

Can you walk me through how to Factor the Polynomial?........54c³d(with exponent of 4) + 9c(exponent of 4)d²

Mathematics
1 answer:
mixas84 [53]3 years ago
8 0

Answer:

I don't know llllllllllllllllllllll

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Consider angle θ in Quadrant II, where sinθ = 4/5 . What is the value of cosθ?
Kay [80]

Answer:

3/5

Step-by-step explanation:

sin theta=opposite/hypotenuse

4/5=opposite/hypotenuse

therefore opposite=4 and hypotenuse=5

for adjacent

using pythagoras theorem

a^2+b^2=c^2

opposite^2 + adjacent^2 =hypotenuse^2

4^2 + adjacent^2 =5^2

16 + adjacent^2 =25

adjacent^2 =25-16

adjacent =\sqrt{9

adjacent=3

cos theta=adjacent/hypotenuse

=3/5

therefore the value of cos theta is 3/5

8 0
3 years ago
Solve 7x-9=28+4(x-1)
ANEK [815]
I got 10.333333333333
8 0
4 years ago
Why doesn’t inverse variation have no x and y intercept?
kirill115 [55]

Answer:

Since k is constant, we can find k given any point by multiplying the x-coordinate by the y-coordinate. For example, if y varies inversely as x, and x = 5 when y = 2, then the constant of variation is k = xy = 5(2) = 10. Thus, the equation describing this inverse variation is xy = 10 or y = .

I hope this was the answer u were looking for.

6 0
3 years ago
Kind of confused on this question, please help
Wittaler [7]

Answer:

4th option

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = \frac{3}{5} x - 6 ← is in slope- intercept form

with slope m = \frac{3}{5}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{3}{5} } = - \frac{5}{3}

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Here m = - \frac{5}{3} and (a, b ) = (- 2, 5 ) , then

y - 5 = - \frac{5}{3} (x - (- 2) ) , that is

y - 5 = - \frac{5}{3} (x + 2)

8 0
3 years ago
PLEASE HELP ASAP!!
murzikaleks [220]

Answer:

Hi,

Step-by-step explanation:

x>=0

y>=0

x+y-5<=0 line passing through (0,5)  and (5,0) with (0,0) in minus region

2x-y-8<=0 line passing trough (0,-8) and (4,0) with (0,0) in minus region

Answer C

6 0
2 years ago
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