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3 years ago

Some please help me I’ll give you brainlist answer

Mathematics

2 answers:

nika2105 [10]3 years ago

5 0

Answer:

The area of the square is 9 aka $3^{2}$

You can find it easily by just counting the number of squares inside the square, or by counting the number of squares at its length then multiplying it by its width; so 3 x 3 = 9

It wouldn't be as easy to find the area of the circle because you don't have the radius, nor diameter, so basically, no quantitative values could be used to at least guess.

Montano1993 [528]3 years ago

3 0

Answer:

Step-by-step explanation:

The area of the square is 3 units by 3 units. because the area of a square is side length squared.

side length = 3

side length^2 = 3 * 3 = 9

The area of a circle is much harder to reason out. You could start with the area of a square and use the diameter is 3 units. But the corners are rounded and you don't know off hand what the area of one of them is, let alone all 4. You are given a formula for circles area, but you have no real idea what the formula actually means. Does it take the corner area into account? At the beginning levels, we don't know.

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X=? <br> Enter the number that goes in the green box.<br> Please helpppp:(

Sergio [31]

Answer:

x=35°

Step-by-step explanation:

145°+110°+70°=325°

360°-325°=35°

4 0

3 years ago

Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

8 0

2 years ago

What adds up to 7 and can also be multiplied to -30

faust18 [17]

X + y = 7
xy = -30

x + y = 7
x - x + y = -x + 7
y = -x + 7

  xy = -30
x(-x + 7) = -30
x(-x) + x(7) = 30
-x² + 7x = -30
-x² + 7x + 30 = 0
-1(x²) - 1(-7x) - 1(-30) = 0
  -1(x² - 7x - 30) = 0
-1 -1
  x² - 7x - 30 = 0
  x² - 10x + 3x - 30 = 0
x(x) - x(10) + 3(x) - 3(10) = 0
  x(x - 10) + 3(x - 10) = 0
(x + 3)(x - 10) = 0
  x + 3 = 0 or x - 10 = 0
- 3 - 3 + 10 + 10
x = -3    or    x = 10

x + y = 7    or   x + y = 7
  -3 + y = 7 or 10 + y = 7
+ 3 + 3 - 10 - 10
y = 10     or y = -3
(x, y) = (-3, 10) or (x, y) = (10, -3)

The two numbers that add up to 7 and multiply to 30 are -3 and 10.

7 0

4 years ago

I Will Give Brainliest, Need Help ASAP

posledela

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A new centrifugal pump is being considered for an application involving the pumping of ammonia. The specification is that the fl

ollegr [7]

Answer:

from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

Step-by-step explanation:

Given that;

$x^{bar}$ = 6.5 gpm

μ = 5 gpm

n = eight runs = 8

standard deviation σ = 1.9 gpm

Test statistics;

t = ( $x^{bar}$ - μ) / $\frac{s}{\sqrt{n} }$

we substitute

t = (6.5 - 5) / $\frac{1.9}{\sqrt{8} }$

t = 1.5 / 0.67175

t = 2.23

the degree of freedom df = n-1 = 8 - 1

df = 7

Now, from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

3 0

3 years ago