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2 years ago

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A plumber needs to replace 20 feet of copper piping. When he gets to the supply store the lengths are given in meters. How many

meters of piping does he need to purchase?

1 answer:

KengaRu [80]2 years ago

3 0

The Meters of piping Needed to be purchased by the plumber are 6.096 meters.
HOPE I HELPED!
-"UNICORN"

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Elsa tries to solve the following equation, and determines there is no solution. Is she correct? Explain.

Vlad [161]

Elsa's answer is incorrect since there is a solution of the given equation. In the given logarithmic problem, we need to simplify the problem by transposing log2(3x+5) in the opposite side. The equation will now be log2x-log2(3x+5)=4. Using properties of logarithm, we further simplify the problem into a new form log (2x/6x+10)=4. Then transform the equation into base form 10^4=(2x/6x+10) and proceed in solving for x value which is equal to 1.667.

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10x^3-35x^2 please help me

marin [14]

Answer:

5x^2(2x-7)

Step-by-step explanation:

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2 years ago

100 points! Will mark brainliest! Only answer if u know it. Random answers will be reported

Setler79 [48]

Step-by-step explanation:

Hey there!

According to the question, you need to find out the distance from your friend's house to school.

<em>Firstly figure out the coordinate of your school and school. Then use formula of distance.</em>

So, we find the coordinates as (7,7) and (5,0).

So, x 1 = 7 y1= 7

x 2 = 5 y2 = 0

~ Use distance formula.

$d = \sqrt{( {x2 - x1)}^{2} + ( {y2 - y1)}^{2} }$

~ Put all values.

$d = \sqrt{ {(5 - 7)}^{2} + ( {0 - 7)}^{2} }$

~ Simplify it.

$d = \sqrt{ {( - 2)}^{2} + {( - 7)}^{2} }$

$d = \sqrt{4 + 49}$

$d = \sqrt{53}$

D= 7.280

Therefore, the distance between your friend's house and your school is 7.3 units.

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)

Tems11 [23]

The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

$2x^2-2=2(x^2-1)=2(x-1)(x+1)$

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

$\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)$

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

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The demand function for a particular product is given by D(x)=−0.75x2+5x+50‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

galben [10]

Step-by-step explanation:

hi nice to meet you my name is anna

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