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2 years ago

Find the 89th term of the arithmetic sequence 25, 35, 45, ...

Mathematics

1 answer:

Eddi Din [679]2 years ago

8 0

Answer:

905

Step-by-step explanation:

arithmetic sequence are in form

a(n)=a(0)+d(n-1)

a(0)=25

common difference: d=35-25=10

a(n)=25+10(n-1)

a(89)=25+10(88)=905

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The reasons as well as the answer

Radda [10]

Answer:

x = 46, y = 67

Step-by-step explanation:

x = ∠ AEC = 46 ( alternate angles )

Since Δ ACE is isosceles then the base angles are congruent, then

y = $\frac{180-46}{2}$ = $\frac{134}{2}$ = 67

4 0

3 years ago

Read 2 more answers

Last week at the park Haeather ran 4 miles in 48 minutes when she runs her next race she Hope's to improve her time by 2 minutes

Korvikt [17]

Answer:

Haeather ran at a speed of 5 miles per hour.

Step-by-step explanation:

Given that last week at the park Haeather ran 4 miles in 48 minutes, to determine what was Haeather rate of speed in miles per hour the following mathematical calculation must be performed:

48/4 = 12

Thus, Haeather ran 1 mile every 12 minutes.

60/12 = 5

1 x 5 = 5

Therefore, Haeather ran at a speed of 5 miles per hour.

3 0

3 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$