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3 years ago

A driveway is in the shape of a rectangle 3.7 meter wide by 4.2 meters long. What is it’s perimeter

Mathematics

2 answers:

Varvara68 [4.7K]3 years ago

8 0

Answer:

15.8

Step-by-step explanation:

since you're trying to figure out the perimeter of the rectangle you would use the formula 2L + 2W.

3.7 times 2 is 7.4

and 4.2 times 2 is 8.4

so, add 7.4 and 8.4

madreJ [45]3 years ago

5 0

Answer:

15.8m

Step-by-step explanation:

Your perimeter is given by the sum of four sides of the rectangle:

Perimeter = 3.7m*2 + 4.2m*2 = 7.4m + 8.4m = 15.8m

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Find all zeros. One zero had been give

nekit [7.7K]

Answer:

The three zeros of the original function f(x) are {-1/2, -3, -5}.

Step-by-step explanation:

"Synthetic division" is the perfect tool for approaching this problem. Long div. would also "work."

Use -5 as the first divisor in synthetic division:

------------------------

-5 2 17 38 15

-10 -35 -15

--------------------------

2 7 3 0

Note that there's no remainder here. That tells us that -5 is indeed a zero of the given function. We can apply synthetic div. again to the remaining three coefficients, as follows:

-------------

-3 2 7 3

-6 -3

-----------------

2 1 0

Note that the '3' in 2 7 3 tells me that -3, 3, -1 or 1 may be an additional zero. As luck would have it, using -3 as a divisor (see above) results in no remainder, confirming that -3 is the second zero of the original function.

That leaves the coefficients 2 1. This corresponds to 2x + 1 = 0, which is easily solved for x:

If 2x + 1 = 0, then 2x = -1, and x = -1/2.

Thus, the three zeros of the original function f(x) are {-1/2, -3, -5}.

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3 years ago

!; im pretty stuck, so which graph would describe JKL and J'K'L' its image after a translation?

Goryan [66]

Answer: A

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the answer to this question 3x+1=10

Angelina_Jolie [31]

Step-by-step explanation:

$3x+1=10 \\ \\ \therefore \: 3x = 10 - 1 \\ \\ \therefore \: 3x = 9 \\ \\ \therefore \: x = \frac{9}{3} \\ \\ \: \: \: \: \: \: \: \: \huge \purple { \boxed{\therefore \: x = 3}} \\ \\$

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2 years ago

What is the solution to he equation (y/y-4)-(4/y+4)=3^2/y^2-16

kicyunya [14]

Answer:

$y=\pm i \sqrt{7}$

Step-by-step explanation:

Given equation is $\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

Factor denominators then solve by making denominators equal

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$\frac{y\left(y+4\right)-4\left(y-4\right)}{\left(y-4\right)\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$y\left(y+4\right)-4\left(y-4\right)=9$

$y^2+4y-4y+16=9$

$y^2=9-16$

$y^2=-7$

take squar root of both sides

$y=\pm \sqrt{-7}$

$y=\pm i \sqrt{7}$

Hence final answer is $y=\pm i \sqrt{7}$ .

7 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

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2 years ago