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3 years ago

I need help in this please help

Mathematics

1 answer:

marta [7]3 years ago

8 0

110 is the answer that I think because if the degree of it is 115 and it is asking for the nearest tenth, then it is 110.

I'm glad to help! :D

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andrezito [222]

Answer: No solution.

Step-by-step explanation:

These lines are parallel to each other. So, they never intersect. So, they don't have solution.

7 0

4 years ago

Can someone help me asap

djyliett [7]

Answer:

1) 4

2) 5

3) -23

4) 8

5) 7

6) -64

7) -10

8) 19

9) 9

10) -56

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3 years ago

Read 2 more answers

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Hatshy [7]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Taylor, Jillian, and Gabby are trying to earn enough money to rent a beach house for a week. They estimate that it will cost at

alekssr [168]

Answer:

the money they need (1650$) minus the money they made (600$)

= 1050

1050 divided by 3 = 350

so they each need to make 350$

Step-by-step explanation:

4 0

3 years ago

Write the first three terms of the series for which tn = 2(n+3). Find the number of terms of the series required to make the sum

ira [324]

Answer:

The first three terms of the series are 8, 10 and 12. The number of terms is 12 to make the sum 228.

Step-by-step explanation:

The series is defined as

$t_n=2(n+3)$

Put n=1.

$t_1=2(1+3)=2\times 4=8$

Put n=2.

$t_2=2(2+3)=2\times 5=10$

Put n=3.

$t_3=2(3+3)=2\times 6=12$

The first three terms of the series are 8, 10 and 12.

It is an arithmetic series. The first terms is 8 and the common difference is

$d=a_2-a_1=10-8=2$

The sum of n terms of an arithmetic series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$288=\frac{n}{2}[2(8)+(n-1)2]$

$288=\frac{2n}{2}[8+n-1]$

$288=n[n+7]$

$0=n^2+7n-288$

$0=n^2+19n-12n-288$

$0=n(n+19)-12(n+19)$

$0=(n+19)(n-12)$

Equate each factor equal to zero.

$n=-19,12$

The number of terms can not be negative, therefore the value of n must be 12.

6 0

3 years ago