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FromTheMoon [43]
3 years ago
10

Pleassee help me asap

Mathematics
1 answer:
earnstyle [38]3 years ago
6 0

Answer:

2 1/2

Step-by-step explanation:

Hope this helps :P

Have a great day!

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What is the equation of the line through the points (3,2), (8,5), (13,8), and (18,11)
olga_2 [115]
For this question the answer is the second one
4 0
3 years ago
Input the equation of the given line in slope intercept form. The line through 1,0 and parallel to x-y=7
Crank
X - y = 7
y = x - 7....the slope here is 1. A parallel line will have the same slope.

y = mx + b
slope(m) = 1
(1,0)...x = 1 and y = 0
now we sub and solve for b, the y int
0 = 1(1) + b
0 = 1 + b
-1 = b

so ur parallel line is : y = x - 1

3 0
4 years ago
Read 2 more answers
The 10th, 4th and 1st term of an A.P are three consecutive find the common ratio of the G.P and the sum of the 6th term, taking
valina [46]

Answer:

S_6 = 252

Step-by-step explanation:

We are told that the 10th, 4th and 1st term of an A.P are three consecutive terms of a G.P.

Now,formula for nth term of an AP is;

a_n = a + (n - 1)d

Thus;

a_10 = a + (10 - 1)d

a_10 = a + 9d

Also;

a_4 = a + (4 - 1)d

a_4 = a + 3d

First term is a.

Thus, since they are consecutive terms of a G.P, it means that;

(a + 9d)/(a + 3d) = (a + 3d)/a

Cross multiply to get;

a(a + 9d) = (a + 3d)(a + 3d)

a² + 9ad = a² + 6ad + 9d²

a² will cancel out to give;

9ad - 6ad = 9d²

3ad = 9d²

Divide both sides by 3d to get;

a = 3d

We are told that the first term is 4.

Thus, 4 = 3d

d = 4/3

We saw earlier that ratio of the GP is (a + 3d)/a

Thus; r = (4 + 3(4/3))/4 = 8/4 = 2

Sum of n terms of a G.P is given by;

S_n = a(rⁿ - 1)/(r - 1)

S_6 = 4(2^(6) - 1)/(2 - 1)

S_6 = 252

8 0
3 years ago
Factor the common factor out of each expression.
jasenka [17]

Answer:

Step-by-step explanation:

10(5k+2)

14) confused sorry

5(a+2)

4(k+3)

5 0
3 years ago
Describe how the cosecant, secant and cotangent are related to the first three trigonometric functions learned.
saw5 [17]
\bf \underline{sin(\theta)}=\cfrac{1}{csc(\theta)}
\qquad 
% cosine
\underline{cos(\theta)}=\cfrac{1}{sec(\theta)}
\qquad 
% tangent
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\\\\\\
% cotangent
csc(\theta)=\cfrac{1}{\underline{sin(\theta)}}
\qquad 
% secant
sec(\theta)=\cfrac{1}{\underline{cos(\theta)}}\qquad 
\begin{array}{llll}
cot(\theta)
\begin{cases}
\cfrac{cos(\theta)}{sin(\theta)}\\\\
\cfrac{1}{tan(\theta )}
\end{cases}
\end{array}
5 0
4 years ago
Read 2 more answers
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