Hi! Can you help me answer this?

Answer:

[tex]V = \frac{4}{3} \pi {r}^{3} \\ \\ 3V = 4\pi {r}^{3} \\ {r}^{3} = \frac{3V}{4\pi} \\ \\ r = \sqrt[3]{ \frac{3V}{4\pi} } [/tex]

The gallons of irrigation water is 326,700 gallons per square foot

Given:

area of the land, A = 43,560 ft²/acre

7.5 gallons = 1 ft³

To find:

number of gallons per square foot

Note:

The number of gallons per square foot is calculated as;

[tex]= \frac{43,560 \ acre}{ft^2} \times \ foot\times \frac{7.5 \ gallons}{ft^3} \\\\= \frac{43,560 \ acre-ft}{ft^2} \times \frac{7.5 \ gallons}{ft^3} \\\\= \frac{43,560 \ ft^3}{ft^2} \times \frac{7.5 \ gallons}{ft^3}\\\\= \frac{43,560 \ ft^3}{ft^3} \times \frac{7.5 \ gallons}{ft^2}\\\\=(43,560\times 7.5) \frac{gallons}{ft^2} \\\\= 326,700 \ \frac{gallons}{ft^2}[/tex]

Therefore, the gallons of irrigation water per square foot is 326,700 gallons per square foot.

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The expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters

From the question, the side lengths are

(4.6x-4.4y)(4.6x−4.4y) cm, (7.5x-8.8z)(7.5x−8.8z) cm, and  (7.7z-9.2y)(7.7z−9.2y) cm.

First, we will clear the brackets one after the other

For  (4.6x-4.4y)(4.6x−4.4y) cm

[tex]4.6x(4.6x-4.4y) -4.4y(4.6x-4.4y)[/tex]

[tex]21.16x^{2} -20.24xy -20.24xy+19.36y^{2}[/tex]

[tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex]

∴ (4.6x-4.4y)(4.6x−4.4y) cm = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm

For  (7.5x-8.8z)(7.5x−8.8z) cm

[tex]7.5x(7.5x-8.8z) -8.8z(7.5x-8.8z)[/tex]

[tex]56.25x^{2} - 66xz -66xz + 77.44z^{2}[/tex]

[tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]

∴ (7.5x-8.8z)(7.5x−8.8z) cm = [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]  cm

For (7.7z-9.2y)(7.7z−9.2y) cm

[tex]7.7z(7.7z-9.2y)-9.2y(7.7z-9.2y)[/tex]

[tex]59.29z^{2} - 70.84yz-70.84yz+84.64y^{2}[/tex]

[tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex]

∴ (7.7z-9.2y)(7.7z−9.2y) cm =  [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm

Now, for the expression that represents the perimeter of the triangle,

Perimeter of a triangle can be calculated by determining the sum of all its sides

That is,

Perimeter of the triangle =  [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm + [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]  cm + [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm

[tex]=21.16x^{2} -40.48xy+19.36y^{2} + 56.25x^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz+84.64y^{2}[/tex]

Collect like terms

[tex]= 21.16x^{2}+ 56.25x^{2} -40.48xy+19.36y^{2}+84.64y^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz[/tex]

[tex]= 77.41x^{2} -40.48xy+104y^{2} - 132xz + 136.73z^{2} - 141.64yz[/tex]

[tex]= 77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm

Hence, the expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters

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Answer:

See attached

Step-by-step explanation:

Answer:

a. 3/4 of pocket money left

b. R10

Step-by-step explanation:

a. 4/4 - 1/4 = 3/4

b. 40/4 = 10 = 1/4 of pocket money

(a). The fraction of pocket money left is 3/4

(b). If he had R40 pocket money, he spend R10.

The ratio is defined as a relationship between two quantities, it is expressed one divided by the other.

Arithmetic operations can also be specified by the subtract, divide, and multiply built-in functions.

The operator that perform arithmetic operation are called arithmetic operators .

Operators which let do basic mathematical calculation

+ Addition operation : Adds values on either side of the operator.

For example 4 + 2 = 6

- Subtraction operation : Subtracts right hand operand from left hand operand.

for example 4 -2 = 2

* Multiplication operation : Multiplies values on either side of the operator

For example 4*2 = 8

Mandla spent one-quarter of his pocket money on sweets

Let x be the total amount of pocket money Mandla had originally

Solution of (a).

⇒ x - (1/4)x

⇒ (3/4)x

The fraction of pocket money left = 3/4

Solution of (b).

1/4 of pocket money

⇒ (1/4)x

⇒ (1/4)40

⇒ 10

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#SPJ2

Answer:

e^{sin²x}+c

Step-by-step explanation:

[tex]\int e^{sin^2x} sin 2x dx=?[/tex]

is this statement?

if so

then

[tex]put~sin^2x=t\\differentiate\\2 sin ~x~cos~x~dx=dt\\sin~2x ~dx=dt\\\int e^t~dt=e^t+c\\=e^{sin^2x}+c[/tex]

Answer:

Step-by-step explanation:

Let's solve your inequality step-by-step.

y - 6 > 2y - 4

y - 2y > -4 + 6

-y > 2

now divide by -1 and inequality sign changes

-y/-1 < 2/-1

y < -2

It would take 17 months and 14 days for the investment to grow to $1545.

To determine the time in months it would take for a $ 1500 dollar investment in a TFSA to grow to $ 1545 if the simple interest at a rate paid was 2% per annum, the following calculation must be performed:

First, you must obtain 2% of 1545 to determine the interest generated per year.

Then, a cross multiplication must be carried out considering the number of months it took to generate said interest, and compare it with the interest that arises from the subtraction of 1545 - 1500, that is, 45.

Therefore, it would take 17 months and 14 days for a $ 1500 dollar investment in a TFSA to grow to $ 1545.

Learn more about interest in https://brainly.com/question/19903178.

Answer:

O (-4,1)

I hope I helped you^_^

Answer:

77/306 or around 25.2%

Step-by-step explanation:

[tex]\frac{7}{18} *\frac{11}{17}[/tex] section 1/total * section 2/(total-1) since there is no replacement

just solve and you get 77/306

9514 1404 393

Answer:

  x = 2 or x = 9

Step-by-step explanation:

To use the quadratic formula, we first need the equation in standard form for a quadratic. We can get there by multiplying the equation by 3(x -3).

  2(3) +4(3(x -3)) = (x +4)(x -3)

  6 +12x -36 = x² +x -12

  x² -11x +18 = 0

Using the quadratic formula to find the solutions, we have ...

  [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-11)\pm\sqrt{(-11)^2-4(1)(18)}}{2(1)}\\\\x=\dfrac{11\pm\sqrt{49}}{2}=\{2,9\}[/tex]

The solutions are x=2 and x=9.

Answer:

(0,0) is the x and y intercept of the function

Step-by-step explanation:

The parabola touches the x and y axis at 0

The x intercept is (0,0) and the y intercept is (0,0)

Answer:

(0, 0) is the x and y intercepts

Step-by-step explanation:

intercepts are where the curve of the equation contacts an axis

The equation is y = x²

Answer:

48

Step-by-step explanation:

Pythagorean Theorem = h^2=p^2+b^2

We have,

(Hypotenuse)h=50

Let 14 be p, i.e (Perpendicular ,Known side)p=14

(Remaining side ,base)b=?(

Now,  

h^2=p^2+b^2

or, 50^2=14^2+b^2

or, 2500-196=b^2

or, √2304=b

b=48

9514 1404 393

Answer:

  9 square inches

Step-by-step explanation:

The area is proportional to the square of the linear scale factor. We can use this to write the proportion ...

  A/(144 m²) = ((2 in)/(8 m))²

  A = (144·4/64) in² = 9 in²

The area representing 144 square meters is 9 square inches.

Answer:

Step-by-step explanation:

#include <stdio.h>

int main()

{

  int num1, num2, flag_var, i, j;

  /* Ask user to input the from/to range

   * like 1 to 100, 10 to 1000 etc.

   */

  printf("Enter two range(input integer numbers only):");

  //Store the range in variables using scanf

  scanf("%d %d", &num1, &num2);

  //Display prime numbers for input range

  printf("Prime numbers from %d and %d are:\n", num1, num2);

  for(i=num1+1; i<num2; ++i)

  {

     flag_var=0;

     for(j=2; j<=i/2; ++j)

     {

        if(i%j==0)

        {

           flag_var=1;

           break;

        }

     }

     if(flag_var==0)

        printf("%d\n",i);

 }

 return 0;

}

9514 1404 393

Answer:

  t = (V -6(3.14)c^2)/(4(3.14)c)

Step-by-step explanation:

Isolate the term containing t, then divide by the coefficient of t

  [tex]V=4(3.14)ct+6(3.14)c^2\\\\V-6(3.14)c^2=t(4(3.14)c)\\\\\boxed{t=\dfrac{V-6(3.14)c^2}{4(3.14)c}}[/tex]

Answer:

0.40905 - 0.10204 = .30701 = 30.7 %

Step-by-step explanation:

0.23 0.40905

1.27 0.10204

Answer:

14m

Step-by-step explanation:

[tex]7m-7+7m-7\\[/tex]

First, we need to eliminate the like term and collect the like term.

[tex]-7+-7=0[/tex]

Now, we have 7m +7m, sum them up and you will get the answer.[tex]7m+7m=14m[/tex]

So, the answer is 14m.

Answer:

14m

Step-by-step explanation:

7m -7 +7m +7​

7m + 7m - 7 + 7

14m

Answer:

Step-by-step explanation: Hello! Do

You're looking for a solution of the form

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]

Differentiating twice yields

[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]

[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]

Substitute these series into the DE:

[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]

[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]

which indicates that the coefficients in the series solution are governed by the recurrence,

[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]

Use the recurrence to get the first few coefficients:

[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]

Just to confirm this solution is valid: we have

y = 10x - 7eˣ   ==>   y (0) = 0 - 7 = -7

y' = 10 - 7eˣ   ==>   y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

Answer:

16

Step-by-step explanation:

Inverse variation is of the form

xy = k where k is a constant

x=4 and y = 32

4*32 = k

128 = k

xy = 128

Let x = 8

8y = 128

Divide each side by 8

8y/8 = 128/8

y =16

Answer:

There are 28 beads

Step-by-step explanation:

Total ratio:

[tex]{ \sf{ (4 + 7) = 11}}[/tex]

let total beads be x:

[tex]{ \sf{ \frac{4}{11} \times x = 16 }} \\ \\ { \sf{x = \frac{11 \times 16}{4} }} \\ x = 44 \: beads[/tex]

Blue beads:

[tex] = 44 - 16 \\ = 28 \: \: beads[/tex]

Answer:

2 is the factor of every even number hope this help you

with no further informations, just go by looking at it.

it's 90°, all other options are too far off

9514 1404 393

Answer:

  the angle is 30°

Step-by-step explanation:

The area of a triangle given two sides and the included angle is ...

  A = 1/2ab·sin(C)

  5 cm² = 1/2(4 cm)(5 cm)sin(θ)

  0.5 = sin(θ) . . . . . . divide by 10 cm²

  θ = arcsin(.5) ≈ 30°

_____

Additional comment

An obtuse angle of 150° in that location will give a triangle with the same area.

Answer:

Roots are -π/2 and π/2

Step-by-step explanation:

[tex]{ \bf{f(x) = 3 \cos(x) }}[/tex]

when x is -2π:

[tex]{ \sf{f( - 2\pi) = 3 \cos( - 2\pi) }} \\ { \sf{ = 3}}[/tex]

hence -2π is not a zero of the function

when x is 2π:

[tex]{ \sf{f(2\pi) = 3 \cos(2\pi) }} \\ { \sf{ = 3}}[/tex]

hence 2π is not a zero of the function

when x is π/2:

[tex]{ \sf{f( \frac{\pi}{2}) = 3 \cos( \frac{\pi}{2} ) }} \\ { \sf{ = 0}}[/tex]

Hence ±π/2 is the zero of the function.