How to solve your problem
(4−3)(−22−7−5)
(4x−3)(−2x2−7x−5)(4x-3)(-2x^{2}-7x-5)(4x−3)(−2x2−7x−5)
Simplify
1
Distribute
(4−3)(−22−7−5)
(4x−3)(−2x2−7x−5){\color{#c92786}{(4x-3)(-2x^{2}-7x-5)}}(4x−3)(−2x2−7x−5)
4(−22−7−5)−3(−22−7−5)
2
distribute
4(−22−7−5)−3(−22−7−5)
4x(−2x2−7x−5)−3(−2x2−7x−5){\color{#c92786}{4x(-2x^{2}-7x-5)}}-3(-2x^{2}-7x-5)4x(−2x2−7x−5)−3(−2x2−7x−5)
−83−282−20−3(−22−7−5)
3
−83−282−20−3(−22−7−5)
−8x3−28x2−20x−3(−2x2−7x−5)-8x^{3}-28x^{2}-20x{\color{#c92786}{-3(-2x^{2}-7x-5)}}−8x3−28x2−20x−3(−2x2−7x−5)
−83−282−20+62+21+15
4
Solution
−83−222++15
I know this looks like a lot but its just how you solve your problem.
therefor, your answer is
Solution
−83−222++15
<em>Hope this helps!</em>
<em>Have a great day!</em>
<em>-Hailey</em>
You need the distributive propery so you do 5 times 3p and get 15p and then 5times 4r and get 20r so 15p plus 20r
Answer: Yes, they both equal -7/2
Step-by-step explanation:
First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
Answer:
An integer is any negative or positive number that’s not a fraction. Ex: 1, -3
Step-by-step explanation:
