Find the distance between the points (0,-9) and(4,-6)

Mathematics

1 answer:

Len [333]3 years ago

8 0

Answer:

d=5

Step-by-step explanation:

1. use the distance formula: d= $\sqrt{(x_{2} - x_{2} )^{2} +(y_{2} -y_{1} )^{2}$

2. plug in the points: d= $\sqrt{(0-4)^{2} +(-9+6)^{2}$

3. solve problem inside parentheses: d= $\sqrt{(-4)^{2} +(-3)^{2}$

4. square -4 and -3: d= $\sqrt{16+9$

5. add:d= $\sqrt{25$

6. find square root: d=5

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One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi

Marrrta [24]

So he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10$

and I'm sure you know what that is

6 0

3 years ago

2x^2 -7= 74 please show all work

skad [1K]

$2x^2-7=74\\
2x^2=81\\
x^2=\dfrac{81}{2}\\
x=-\sqrt{\dfrac{81}{2}} \vee x=\sqrt{\dfrac{81}{2}}\\
x=-\dfrac{9}{\sqrt2} \vee x=\dfrac{9}{\sqrt2}}\\
x=-\dfrac{9\sqrt2}{2} \vee x=\dfrac{9\sqrt2}{2}}\\$