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2 years ago

If you travel at a rate of 45 miles per hour.How many minutes will it take to travel 1 mile

1 answer:

8 0

<span>You can apply a rule of three.
If you travel a distance of 45 miles in one hour (equivalent to 60 minutes), then how many minutes (x) will it take 1 mile?

45 miles → 60 min
1 mile → x

</span> $x=\frac{1 mile*60 min}{45 miles}$ --> $x=\frac{60}{45} min$ <span>

Simplifying the fraction, dividing the numerator and denominator by 15

x=(60/15)/(45/15) min→x=4/3 min =1.333333333 min
approximately 1.33 min

Answer: It will take 4/3 min or 1.33 min to travel 1 mile</span>

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What is the factorization of the trinomial below -x^2+x+42

aksik [14]

For the given trinomial

$a=-1, b=1 ,c=42$

We need to multiply a by c
$-1 \times 42 = - 42$

We split the middle term with these two factors
$7,-6$
To obtain,

$- {x}^{2} + 7x -6x + 42$
We now factor

$- {x}^{2} + x + 42 \: = x( - x + 7) + 6( - x + 7)$
This implies

$- {x}^{2} + x + 42: = (x + 6)(7 - x)$

4 0

3 years ago

Read 2 more answers

Solve by using the quadratic formula.<br> 3psquared+7p+2=0

Rainbow [258]

Answer:

$\large\boxed{x=-2\ or\ x=-\dfrac{1}{3}}$

Step-by-step explanation:

The quadratic formula of a quadratic equation

$ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

We have the equation:

$3p^2+7p+2=0\to a=3,\ b=7,\ c=2$

Substitute:

$b^2-4ac=7^2-4(3)(2)=49-24=25\\\\\sqrt{b^2-4ac}=\sqrt{25}=5\\\\x_1=\dfrac{-7-5}{2(3)}=\dfrac{-12}{6}=-2\\\\x_2=\dfrac{-7+5}{2(3)}=\dfrac{-2}{6}=-\dfrac{1}{3}$

8 0

3 years ago

Read 2 more answers

Factor.<br> 2y? + 13y+21

diamong [38]

Answer:

3(5y+7)

Step-by-step explanation:

4 0

3 years ago

Simplify math please help

jok3333 [9.3K]

Obviously it’s not a answer without x in the numerator so A is not the answer and your not multiplying or adding so your answer is C

4 0

2 years ago

(2+4+6+....+2006)-(1+3+5+....+2005) the ‘....’ means they add up until it gets to 2006 or 2005. Plz tell me how you get the answ

Naya [18.7K]

I will solve the problem by 2 methods.

<h2>Method 1 to solve:</h2>

I will use Gauss's formula for calculating each string, then we will decrease the results.

We find the number of terms of each string.

$\displaystyle\\n_1=\frac{2006-2}{2}+1=\frac{2004}{2}+1=1002+1=1003~~\text{terms}\\ \\n_2=\frac{2005-1}{2}+1=\frac{2004}{2}+1=1002+1=1003~~\text{terms}\\\\n_1=n_2=n=1003~~\text{terms}\\\\\implies~~~\text{The two strings have the same number of terms.}\\\\S_1 =2+4+6+...+2006=\frac{n(2006+2)}{2}=\frac{1003\times2008}{2}=1003\times1004\\ \\S_2=1+3+5+...+2005=\frac{n(2005+1)}{2}=\frac{1003\times2006}{2}=1003\times1003\\\\S_1-S_2=1003\times1004-1003\times1003=1003(1004-1003)=1003\times1=\boxed{1003}$

<h2>Method 2 to solve:</h2>

We intersect the terms in the second string among the terms of the first string.

$\displaystyle\\S=(2+4+6+....+2006)-(1+3+5+....+2005)=\\\\=2+4+6+....+2006-1-3-5-....-2005=\\\\=2-1+4-3+6-5+...+2006-2005=\\\\=\underbrace{(2-1)+(4-3)+(6-5)+...+(2006-2005)}_{1003~~\text{parenthesis.}}=\\\\\text{We know from the first method that each string has 1003 terms}\\\\=\underbrace{1+1+1+....+1}_{\text{1003 terms equal to 1}} = 1003 \times 1 =\boxed{1003}$

3 0

2 years ago