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2 years ago

I have a daily allowance of 70grm but have only used 48 what percentage do i have left

Mathematics

1 answer:

ladessa [460]2 years ago

4 0

Answer:

You have 31.43% of your allowance left.

Step-by-step explanation:

This question can be solved using a rule of three.

Your initial amount, of 70, is 100%.

The remaining amount, of 70 - 48 = 22, is x%. So

70 - 100%

22 - x%

$70x = 100*22$

$x = \frac{100*22}{70}$

$x = 31.43$

You have 31.43% of your allowance left.

You might be interested in

Which math expression means "the product of 16 and 26"? 16 +26 O 16.26 15 - 26 26-16

Llana [10]

Answer:

The product of 16 and 26 would be 16 * 26

Step-by-step explanation:

This is because "the product of" means multiplication so *. Making it 16 * 26.

8 0

3 years ago

Evaluate the following pic

GREYUIT [131]

Answer:

1) $\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}=-11$

3) Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

Step-by-step explanation:

1) $\sqrt{1225}+\sqrt{1024}$

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

$\sqrt{1225}+\sqrt{1024}\\=\sqrt{5\times5\times7\times7}+\sqrt{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}\\=\sqrt{5^2\times7^2}+\sqrt{2^2\times2^2\times2^2\times2^2\times2^2}\\=5\times7+(2\times2\times2\times2\times2)\\=35+32\\=67$

$\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}$

We know that $\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)$

Applying radical rule:

$\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11$

So, $\sqrt[3]{-1331}=-11$

3) $2:p :: p:8$

It can be written as:

$p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4$

Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6$

Put value of x, y and z in equation and solve:

$x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25$

So, $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}$

We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

$\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6$

So, $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

8 0

2 years ago

300€ durante 5 años al 0.3% de interés compuesto mensual. A) ¿Cuánto interés se ha ganado? B) Cuánto interés se ha ganado?

marysya [2.9K]

Answer:

Interes ganado= €59.07

Step-by-step explanation:

Dada la siguiente informacion:

300€ durante 5 años al 0.3% de interés compuesto mensual.

Para calcular el valor final de la inversión y el interés ganado, debemos usar la siguiente formula:

FV= PV*(1 + i)^n

Present Value= 300

i= 0.003

n= 5*12= 60 meses

FV= 300*(1.003^60)

FV= 359.07

Interes ganado= 359.07 - 300

Interes ganado= 59.07

5 0

2 years ago

If sin y° = 7 over q and tan y° = 7 over r, what is the value of sec y°?

Tomtit [17]

Answer:

Sec y = q/r

Step-by-step explanation:

Sin y = 7/q

tan y = 7/r

Mathematically cos y = sin y/tan y

But sec y = 1/cos y

So therefore sec y = tan y/sin y

So that would be ;

7/r divided by 7/q

= 7/r * q/7

= q/r

3 0

3 years ago

Use simpson's rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. compare y

Lubov Fominskaja [6]

y = ln x , 1 <= x <= 3, about x axis and n = 10, dy/dx = 1/ x

S = (b a) ∫ 2π y √( 1 + (dy/dx) ^2) dx

so our f(x) is 2π y √( 1 + (dy/dx) ^2)

(b - a) / n = / 3 = (3-1) / 30 = 1/15

x0 = 1 , x1 = 1.2, x2 = 1.4, x3 = 1.6 ....... x(10) = 3

So we have , using Simpsons rule:-

S10 = (1/15) ( f(x0) + 4 f(x1) + 2 f)x2) +.... + f(x10) )

= (1/15) f(1) + f(3) + 4(f(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)) + 2(f(1.4) + f(1.8) + f(2.2) + f(2.6) )

( Note f(1) = 2 * π * ln 1 * √(1 + (1/1)^2) = 0 and f(3) = 2π ln3√(1+(1/3^2) = 7,276)

so we have S(10)

= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +

5 0

3 years ago