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3 years ago

the width of a rectangle is 6ft. find the perimeter of the rectangle if it's area is 42 square feet.

2 answers:

5 0

A=l x w
substitute
42=l x 6
7=l
Perimeter of a rectangle is 2l +2w
2(6)+ 2(7)
12+14=26ft is the perimeter

Hope this helps

katrin2010 [14]3 years ago

4 0

Formula: Area = length * width
Substitute in what you know: 42 = length * 6
Divide both sides by 6 to get the length: length = 7

Formula: Perimeter = 2 (length + width)
Substitute in what you know: P= 2 (7+6)
Solve: Perimeter = 26 ft

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33. The perimeter of a room is 22m.lf<br> the length of the room is 6.5m,<br> what is its width?

vredina [299]

The formula for perimeter is 2(Length) + 2(width)

Fill in the known information:

22 = 2(6.5)+ 2(width)

Simplify:

22 = 13 + 2(width)

Subtract 13 from both sides:

9 = 2(width)

Divide both sides by 2:

Width = 4.5m

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3 years ago

What are the solutions to log subscript 6 baseline (x squared 8) = 1 log subscript 6 baseline (x)

n200080 [17]

The solution of the logarithmic equation is given by the antilog. Then the value of x is 2 and 4.

<h3>What is a factorization?</h3>

It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.

The logarithmic equation is given below.

$\log _6(x^{2} +8)= 1+ \log _6x\\\\\log _6(x^{2} +8)= \log_66+ \log _6x\\\\\log _6(x^{2} +8)= \log _66x\\$

Take antilog, then we have

$\begin{aligned} x^{2} +8 &= 6x\\\\x^2 -6x + 8 &= 0\\\\x^2 -4x - 2x +8 &=0\\\\(x-4)(x-2) &= 0\\\\x&= 2, 4\end{aligned}$

Then the value of the x is 2 and 4.

More about the factorization link is given below.

brainly.com/question/6810544

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2 years ago

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi

lara [203]

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$ , and evaluate it for the given point $(x,y)$ . This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$ . For example: