All categories

3 years ago

Ben drove 350 miles in 6 1/2 hours Marcie drove 520 miles on 12 1/4 hours who drove faster

Mathematics

1 answer:

natka813 [3]3 years ago

8 0

Answer:

Ben drove faster

Step-by-step explanation:

350 miles / 6.5 hours = 53.84 miles per hour

<h2>MARCIE:</h2>

520 miles / 12.25 hours = 42.44 miles per hour

53.84 is faster than 42.44 therefore Ben is faster.

You might be interested in

HELP! What is the solution to the following system of equations?

tatuchka [14]

Answer:

it is 2,4

Step-by-step explanation:

The point that they meet is 2,4 . It means X+ is 2 and Y+ is 4 .

You can write a as their meet point . you should answer like this a(2,4) .

4 0

3 years ago

adidas is ready to launch its new shoe line, but needs to determine what color the shoes should be so that customers will actual

Dmitry [639]

Answer:

Yellow and purple

Step-by-step explanation:

7 0

4 years ago

Slope-intercept from by steps

nordsb [41]

Y=m+b use this equation

3 0

3 years ago

Factors of 64a3 - 343b 3 are

bearhunter [10]

$\huge\mathfrak\green{Constant}$

Constant is the number that cannot change the value

Im not sure..but you can copy it

8 0

3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

3 years ago