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Norma-Jean [14]
3 years ago
13

#3) PLEASE HELP WITH QUESTION, MARKING BRAINLIEST + POINTS :)

Mathematics
1 answer:
Taya2010 [7]3 years ago
3 0
\boxed { \boxed { a_n = a_1 + d(n - 1)}}

a_{14} = -33 , a_{15} = 9,

a_{14} =  a_1 + d(14 - 1) = -33
a_1 + 13d = -33

a_{15} = a_1 + d(15 - 1) = 9
a_1 + 14d = 9

So the two equations:
a_1 + 13d = -33 ----------------- ( 1 )
a_1 + 14d = 9     ----------------- ( 2 )

(2) - (1) :
d = 42  ----------------- (Sub into 1)

a_1 + 13(42) = -33
a_1 + 546= -33
a_1 = -33 - 546
a_1 = -579

\boxed { \boxed { a_n = -578 + 42(n - 1)}}




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Someone help plz no web sights and plz make sure it’s right :) if it is will mark brainiest
alisha [4.7K]

Cosine is the ratio of the adjacent leg to the hypotenuse. The leg adjacent to A is 30, and the hypotenuse is 50.

The ratio is 30/50, which simplifies to 3/5.

Let me know if you need any clarifications, thanks!

8 0
3 years ago
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Which equation represents a line that passes through (2, –) and has a slope of 3?
Radda [10]
The complete question in the attached figure

Let
(2, -R)----------> the point (2, -?)

the equation of a line is
y-y1=m*(x-x1)
point (2.-R)  and m=3
so

y+R=3*(x-2)

the answer is the option 
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4 0
3 years ago
3 (3x+5)=2 (2x-5) solve for x
marin [14]

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-5

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3 0
2 years ago
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What is the value of 1/3 x^2 + 5.2y when x = 3 and y = 2?<br><br> PLZ HELP!!!
olasank [31]
The first step in solving this is to substitute x with 3 and y with 2, so that
\frac{1}{3} x^{2} +5.2y
becomes
\frac{1}{3}  (3)^{2} +5.2(2)
Then simply solve using the order of operations.
First exponents (there are no parentheses)
\frac{1}{3} 9 +5.2(2)
Then multiply and divide
3+10.4
And finally add and subtract
13.4
Therefore the value <span>of 1/3 x^2 + 5.2y when x=3 and y=2 is 13.4</span>
6 0
3 years ago
an inlet pipe and a hose together can fill the pond in 10 hours. The inlet pipe alone can complete the job in one hour less time
Annette [7]

Answer:

The hose pipe can fill the pond in 21.465 hours.

The inlet pipe can fill the pond in 20.465 hours.

Step-by-step explanation:

Given that, a inlet pipe and a hose pipe together can fill the pond in 10 hours.

The inlet pipe alone can fill the pond in one hour less time than the hose pipe can  fill the pond.

Let the hose pipe can complete the job in x hours.

Then the inlet pipe can fill the pond in (x-1) hours.

The rate of filling of hose pipe is =\frac{1}{x}

The rate of filling of inlet pipe is =\frac{1}{x-1}

The rate of filling of both pipes is = \frac 1{10}

According to the problem,

\frac{1}{x}+\frac{1}{x-1}=\frac{1}{10}

\Rightarrow \frac{x-1+x}{x(x-1)}=\frac1{10}                              [ simplifying the fraction]

\Rightarrow \frac{2x-1}{x^2-x}=\frac1{10}                            

\Rightarrow 10(2x-1)= x^2-x                [ Multiplying 10(x²-x) both sides]

\Rightarrow x^2-x= 20x-10

\Rightarrow x^2-x-20x+10=0             [ simplifying ]

\Rightarrow x^2-21x+10=0

\Rightarrow x=\frac{-(-21)\pm\sqrt{(-21)^2-4.1.10}}{2.1}       [ applying quadratic formula]

\Rightarrow x=\frac{21\pm21.93}{2.1}

⇒ x= 21.465,  -0.465

x= - 0.465 does not possible, since time can not be negative.

∴x=21.465

The hose pipe can fill the pond in 21.465 hours.

The inlet pipe can fill the pond in (21.465-1)=20.465 hours.

3 0
3 years ago
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