All categories

3 years ago

Plspls help!

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

4 0

Answer:

a)=300

b)=79

Step-by-step explanation:

a) 3m²

=3(10)²

=300

b) n³/3 + 7m³/100

=3³/3 + 7(10)³

=9+70

=79

You might be interested in

18.79+2.11+(-1.92)+17.28

stellarik [79]

Answer:

36.26

Step-by-step explanation:

Hope this helps. Plz give brainliest.

7 0

3 years ago

37. Rectangle ABCD is graphed on a coordinate

Firlakuza [10]

AC # A'B'

Step-by-step explanation:

While solving the above problem one must keep in mind that

Reflection along Y-axis would just make the rectangle upside down at the same distance from X-axis in the down direction.

Thus, from the figure it becomes clear that

AB ∦ B’D’
CD║A’B’
BD ∦ C’D’

Only statement that comes out true is the third statement which states that AC is not parallel to A’B’

Since the diagonal of a rectangle can not be parallel to one of its side (though the side in question is side of the same rectangle reflected in y axis)

6 0

3 years ago

How can 5 people share 4 cookies equally?

Volgvan

By giving each person a fraction of a cookie.

4÷5 = 0.80

If ya give 'em 0.80 th of a cookie, it'll be equal!

Thank you ;)

4 0

3 years ago

If a slice of pizza is 2 dollars and I have 15 dollars how many pieces of pizza can I get

Elanso [62]

7 slices of pizza and you will have 1 dollar left over

4 0

2 years ago

Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago