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3 years ago

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Mathematics

2 answers:

ivann1987 [24]3 years ago

8 0

Answer:

I think the answer is B) 4.7

Step-by-step explanation:

just to let you know I maybe wrong.

Sophie [7]3 years ago

3 0

Answer:

4.4

Step-by-step explanation:

4.4

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I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

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3 years ago

On a coordinate plane, 2 triangles are shown. The first triangle has points M (negative 5, 4), N (negative 2, 3), and L (negativ

worty [1.4K]

Answer:

The rule of the reflection is rx-axis(x, y) → (x, –y) ⇒ 3rd answer

Step-by-step explanation:

Let us revise the reflection on the axes

If point (x , y) reflected across the x-axis , then its image is (x , -y) , the rule of reflection is rx-axis (x , y) → (x , -y)

If point (x , y) reflected across the y-axis , then its image is (-x , y) , the rule of reflection is ry-axis (x , y) → (-x , y)

In Δ LMN

∵ L = (-4 , 2)

∵ M = (-5 , 4)

∵ N = (-2 , 3)

In ΔL'M'N'

∵ L' = (-4 , -2)

∵ M' = (-5 , -4)

∵ N' = (-2 , -3)

The signs of y-coordinates of the vertices of Δ LMN are changed, that means Δ LMN are reflected across the x-axis

∵ The y-coordinate of L is 2 and the y-coordinate of L' is -2

∵ The y-coordinate of M is 4 and the y-coordinate of M' is -4

∵ The y-coordinate of N is 3 and the y-coordinate of N' is -3

∴ Δ LMN is reflected across the x-axis

∴ The image of point (x , y) is (x , -y)

The rule of the reflection is rx-axis(x, y) → (x, –y)

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Anarel [89]

Answer:

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Step-by-step explanation:

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