I need the find the area of a rectangle. the length is 6.5x + 5ft and the width is 15 ft

PLEASE HELP ME FAST! WILL GIVE BRAINLIEST.

soldier1979 [14.2K]

Answer- √ 26

Explanation:

Use the distance formula:

√ ( y 2 − y 1 )² + ( x 2 − x 1 ) ²

Plug in your values:

√ ( − 5− ( − 4 ) )² + ( 2 − ( − 3 ) )²

Simplify:

√ ( − 1 )² + ( 5 )²

Simplify:

√ 1 + 25

Simplify:

√ 26

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Anyways bye!

8 0

3 years ago

Write the decimal place name for each of the underlined digits. a. .423 b. .32678 c. .4524 d. .3006

kogti [31]

A= 4 tenth 2 hundredth 3 thousandth
b= 3 tenth 2 hundredth 6 thousandth 7 ten thousandth 8 hundred thousandth
c= 4 tenth 5 hundredth 2 thousandth 4 ten thousandth
d= 3 tenth 0 hundredth 0 thousandth 6 ten thousandth

5 0

4 years ago

Read 2 more answers

PLZ HELP (will give brainlest) i think thats how u spell it idk

timama [110]

Answer:

Y are you making us do the test question you should have been payin attention in class

Step-by-step explanation:

7 0

3 years ago

Mr.Anderson paid $19.18 for a CD cover. This amount includes a tax of 8.5%. What was the cost of the CD cover before tax ?

MrRa [10]

X * 1.085 = 19.18
solve for x

Why is this how you figure it out? well... you have the price of the CD cover and the total is the price of the CD cover with tax.

Price of the CD cover would be: x * 1 = price of cd cover
Put that tax on there: x * 1.085 = price of cd cover with tax

why .085? move that decimal over two places to the left from the 8.5%

3 0

3 years ago

HALLP QUICKKKK

Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
 $S_{a}=S_{l}-2$
We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so:
 $t_{l}=t_{a}+1$

Now, we can set up our equations.
Speed of the boat traveling in the river:
 $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

3 0

4 years ago