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2 years ago

May i get some help ? i will mark brainliestinclude steps

Mathematics

2 answers:

Julli [10]2 years ago

8 0

Answer:

4.9

Step-by-step explanation:

nya

steposvetlana [31]2 years ago

7 0

Answer:

4.9 m²

Step-by-step explanation:

Area of a Triangle = (base * height)/2

(3.5*1.4)/2 = 2.45 m² = A of CBD

(3.5*1.4)/2 = 2.45 m² = A of BCA

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If C=r+7 and D=-2r+3, find an expression that equals 2C−D in standard form.

quester [9]

4r+11 is the answer

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2 years ago

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Work out the circumference of this circle.

tigry1 [53]

C=pid
Since d=9m and pi=3.142, C=3.142*9m=28.278
Rounded to 1 decimal point =28.3

3 0

3 years ago

Give the positive solution of x^2 – 36 = 5x.

andriy [413]

The equation given in the question is

x^2 - 36 = 5x
x^2 - 5x - 36 = 0
x^2 - 9x + 4x - 36 = 0
x(x - 9) + 4(x - 9) = 0
(x - 9) (x + 4) = 0
From the above set we can say that x = 9 is the positive solution of the equation given in the question. I hope that this is the answer that you were looking for and it has come to your great help.

5 0

2 years ago

What is the slope of the line parallel to 9/10x +4y =70

butalik [34]

Answer:

y=−9/40x+3/52

Using the slope-intercept form, the slope is −9/40.

the slope: m=−9/40

Step-by-step explanation:

3 0

3 years ago

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

6 0

3 years ago