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ch4aika [34]
3 years ago
14

What is the distance between these two points

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

pretty sure it is 11

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15 divided by 3.15 what is the unit rate
stiv31 [10]

Answer: 4.8 i think

Step-by-step explanation: its just division

7 0
3 years ago
Solve the equation x+0.7=1-0.2x In two different ways
timama [110]
Add 0.2x to both sides
1.2x+0.7= 1
1.2x=.3
x=.3/1.2
x=.25


option #2
subtract .7 on both sides

x=1-.7-.2x
x=.3-.2x
1.2x=.3
x=.25
3 0
3 years ago
Find the equation of a line that passes through (3,12) and is parallel to the graph of y=2x+4. Write the equation in​ slope-inte
anygoal [31]

Answer:

12+3x=y

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Prove: cot(x) sec^4(x) = cot(x) +2tan(x) + tan^3(x)
masha68 [24]
\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad 
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad  1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\
-------------------------------\\\\
cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
-------------------------------\\\\

\bf \textit{so, let's do the left-hand-side}\\\\
cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)]
\\\\\\
cot(x)[1^2+2tan^2(x)+tan^4(x)]
\\\\\\
cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x)
\\\\\\
cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)}
\\\\\\
cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}
8 0
3 years ago
What is the next number in the following series: 54, 49, 44, 39, ...
Damm [24]

<em>Hello there! And thank you for asking your question here on brainly . com! The number one asking and answering site.</em>

Answer: C. 34

Explanation: The series shown is reducing by 5 digits. 54 - 5 = 49, 49 - 5 = 44, and so on. When we're at 39, just subtract by 5. 39 - 5 = 34. 34 is the answer.

Hope this helped you, and have an awesome day! ☺♥

7 0
3 years ago
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