I need help ASAP!!!!PLS

Jose just ran a 5 kilometer race. How many meters did he run?

love history [14]

Answer:

A). 5,000 meters

Step-by-step explanation:

value of kilometer is in thousands.

that means:

5×1000= 5000

Jose ran 5000 meters

8 0

3 years ago

Read 2 more answers

I have some answers the these problems but I just wanna make sure there right.. PLEASE Help??

yanalaym [24]

7x+12+63=180
8x+12=180
Subtract 12 from both sides
8x= 168
=c=21

10x-40=50
Add 40 to both sides
10x=90
X=9

4x+72+2x-12=180
6x +60 =180
Subtract 60 from both sides
6x=120
X=20

3 0

3 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

2 years ago

PLEASE, I NEED HELP, ILL GIVE BRAINLY...What will be the perimeter and the area of the rectangle below if it is enlarged using a

lisabon 2012 [21]

(Sorry if I'm wrong)

Enlarged means making it bigger, so you're making it 3 1/2 times bigger.

Try multiplying both the area and perimeter by 3.5

(6 x 3.5) (8 x 3.5)

P = 21cm and A = 28cm

7 0

3 years ago

On a trip to the movies a family spent 35 on tickets and another 25 on popcorn. They alsp spent 10 on drinks. The oldest son spe

rodikova [14]

$6 dollars on candy because 35+25+10=70, 76-70=$6

5 0

3 years ago