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3 years ago

What is the fourth term of the arithmetic sequence described by the explicit formula an = 2 + (n - 1)25?

Mathematics

1 answer:

Elis [28]3 years ago

3 0

Answer:

Step-by-step explanation:

plug in 3 if we're going by 0,1,2,3

2+ (3-1)25

2+50

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A. find the value of a. B. Find the value of the marked angles.

Marat540 [252]

The marked angles are opposite angles, and as such they have the same measure. So, we have

$6a+11 = 2a+83$

Subtract 2a and 11 from both sides to get

$4a = 72$

Divide both sides by 4 to get

$a = \dfrac{72}{4} = 18$

Now that we know the value of a, we can compute the measure of the angles. We can also verify that the solution we found is correct by verifying that both expressions actually give the same result:

$6a+11 = 6\cdot 18 + 11 = 119$

$2a+83 = 2\cdot 18 + 83 = 119$

8 0

3 years ago

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What are the coordinates of Y(-35, 5) is reflected across the x-axis

puteri [66]

Answer:

(- 35, - 5 )

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y ) → (x, - y ), hence

(- 35, 5 ) → (- 35, - 5 )

7 0

3 years ago

Let f(x) = x^2 − 5. Find f(-3)

Art [367]

Answer:

f(-3)=4

Step-by-step explanation:

Substitute x=-3 into the function

f(-3)=(-3)^2-5

f(-3)=9-5

f(-3)=4

5 0

2 years ago

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If x^2=y^2+z^2 what does x equal?

Zepler [3.9K]

Answer:

$\displaystyle x = \sqrt{y^2 + z^2}$

General Formulas and Concepts:

Pre-Algebra

Equality Property

Algebra i

Terms/Coefficients

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle x^2 = y^2 + z^2$

Step 2: Solve for x

[Equality Property] Square root both sides: $\displaystyle x = \sqrt{y^2 + z^2}$

3 0

3 years ago

Find the simplified product b-5/2b x b^2+3b/b-5

White raven [17]

Answer:

The product $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}$

Step-by-step explanation:

Given expression $\frac{b-5}{2b}$ and $\frac{b^2+3b}{b-5}$

We have to find the product of $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}$

Consider the given expression $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}$

Multiply fractions, we have,

$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}$

$=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}$

Cancel common factor ( b - 5 )

we have, $=\frac{b^2+3b}{2b}$

Apply exponent rule,

$\:a^{b+c}=a^ba^c$

$b^2=bb$

$=bb+3b=b(b+3)$

$=\frac{b\left(b+3\right)}{2b}$

Cancel common factor b , we have,

$=\frac{b+3}{2}$

Thus, the product $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}$

8 0

3 years ago

Read 2 more answers