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3 years ago

Help me with number 13 please

Mathematics

1 answer:

Colt1911 [192]3 years ago

6 0

R (-6, 3)
S (-4, 6)
T (1, 4)
U (-4, -1)

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A cyclist travels 20 miles in the same amount of time it takes a hiker to walk 6 miles. The rate of the cyclist is 7 mph faster

Anna11 [10]

Answer:

The speed of the cyclist is 10 miles per hour .

Step-by-step explanation:

Given as :

The distance cover by cyclist = $D_1$ = 20 miles

The distance cover by biker = $D_2$ = 6 miles

The speed of biker = $S_2$ = s mph

The speed of cyclist = $S_1$ = 7 + s mph

The time taken by both cyclist and biker = t hour

Now. Distance = Speed × Time

So, $D_1$ = $S_1$ × t

Or, 20 = s × t

Or, t = $\frac{20}{7+s}$ .....1

And ,

$D_2$ = $S_2$ × t

or, 6 = s t

Or, t = $\frac{6}{s}$ ....2

From Eq 1 and 2

I.e $\frac{20}{7+s}$ = $\frac{6}{s}$

Or, 20 s = 6 ( 7 + s )

Or, 20 s - 6 s = 42

Or, 14 s = 42

∴ s = $\frac{42}{14}$

I,e s = 3 miles per hour

So, The speed of biker = $S_2$ = s = 3 miles per hour

And The speed of cyclist = $S_1$ = 7 + s = 7 + 3= 10 miles per hour

Hence The speed of the cyclist is 10 miles per hour . Answer

4 0

3 years ago

A sprinter can run 279 feet in 9 seconds. Find the sprinter's unit rate of feet per second.

TiliK225 [7]

Easy 31 feet per second take 279 and divide it by 9 and you get 31

5 0

3 years ago

Please help me with this

Gnoma [55]

Given:

The equation is:

$3^x\times 9=3^{2n-1}$

To find:

The value of n in terms of x.

Solution:

We have,

$3^x\times 9=3^{2n-1}$

It can be written as

$3^x\times 3^2=3^{2n-1}$

$3^{x+2}=3^{2n-1}$ $[\because a^ma^n=a^{m+n}]$

On comparing the exponents, we get

$x+2=2n-1$

Add 1 on both sides.

$x+2+1=2n-1+1$

$x+3=2n$

Divide both sides by 2.

$\dfrac{x+3}{2}=n$

Therefore, the value of n is terms of x is $n=\dfrac{x+3}{2}$ .

7 0

3 years ago

Please help me solve this question thank you

GarryVolchara [31]

Answer:

Step-by-step explanation:

7 0

3 years ago

Divide. use patterns and place value. 440 ÷4=

Y_Kistochka [10]

Do 110 times 40 to get 440 you are welcome hope u are good

4 0

3 years ago

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