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Bad White [126]
2 years ago
13

John has 175cm and 1.7m of rope whose rope is longest

Mathematics
1 answer:
Arada [10]2 years ago
6 0

175cm is the longest
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f(x)={x+1]^2 Determine for each x-value whether it is in the domain of f or not. (-2 y/n} { -1 y/n} {9 y/n}
kvv77 [185]

Answer:

  all are "yes"

Step-by-step explanation:

A polynomial is defined for all values of x. None are excluded. Every value listed is in the <em>domain</em> of f(x) = (x +1)².

4 0
3 years ago
Find the value of x when 6 - 4x = 6x - 8x + 12
seraphim [82]

Answer:

x = -3

Step-by-step explanation:

Step 1: Write out equation

6 - 4x = 6x - 8x + 12

Step 2: Combine like terms (x)

6 - 4x = -2x + 12

Step 3: Add 4x on both sides

6 = 2x + 12

Step 4: Subtract 12 on both sides

-6 = 2x

Step 5: Divide both sides by 2

-3 = x

Step 6: Rewrite

x = -3

8 0
2 years ago
Read 2 more answers
Explain how you would find the slope of the line y = 4.
Slav-nsk [51]

Answer:

The slope is 0.

Step-by-step explanation:

You could draw it on graphs paper . It will be a horizontal line passing through the point (0,4). y will always be 4 and x can be any value.

The slope of a horizontal line is 0.

4 0
2 years ago
The arch beneath a bridge is​ semi-elliptical, a​ one-way roadway passes under the arch. The width of the roadway is 38 feet and
forsale [732]

Answer:

Only truck 1 can pass under the bridge.

Step-by-step explanation:

So, first of all, we must do a drawing of what the situation looks like (see attached picture).

Next, we can take the general equation of an ellipse that is centered at the origin, which is the following:

\frac{x^2}{a^2}+\frac{y^2}{b^2}

where:

a= wider side of the ellipse

b= shorter side of the ellipse

in this case:

a=\frac{38}{2}=19ft

and

b=12ft

so we can go ahead and plug this data into the ellipse formula:

\frac{x^2}{(19)^2}+\frac{y^2}{(12)^2}

and we can simplify the equation, so we get:

\frac{x^2}{361}+\frac{y^2}{144}

So, we need to know if either truk will pass under the bridge, so we will match the center of the bridge with the center of each truck and see if the height of the bridge is enough for either to pass.

in order to do this let's solve the equation for y:

\frac{y^{2}}{144}=1-\frac{x^{2}}{361}

y^{2}=144(1-\frac{x^{2}}{361})

we can add everything inside parenthesis so we get:

y^{2}=144(\frac{361-x^{2}}{361})

and take the square root on both sides, so we get:

y=\sqrt{144(\frac{361-x^{2}}{361})}

and we can simplify this so we get:

y=\frac{12}{19}\sqrt{361-x^{2}}

and now we can evaluate this equation for x=4 (half the width of the trucks) so:

y=\frac{12}{19}\sqrt{361-(8)^{2}}

y=11.73ft

this means that for the trucks to pass under the bridge they must have a maximum height of 11.73ft, therefore only truck 1 is able to pass under the bridge since truck 2 is too high.

5 0
2 years ago
Rewrite the expression with a perfect square: x^2 - 14x + 11
nikitadnepr [17]

   Factoring  x2-14x+11 

The first term is,  x2  its coefficient is  1 .

The middle term is,  -14x  its coefficient is  -14 .

The last term, "the constant", is  +11 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 11 = 11 

Step-2 : Find two factors of  11  whose sum equals the coefficient of the middle term, which is   -14 .

     -11   +   -1   =   -12     -1   +   -11   =   -12     1   +   11   =   12     11   +   1   =   12

Observation : No two such factors can be found

4 0
2 years ago
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