Find the H.C.F of 36 and 54

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your questions:

1.

a) Simplify $\log 9 + \log 8$ :

Use the logarithm property $\log a + \log b = \log (a \times b)$ .
$\log 9 + \log 8 = \log (9 \times 8)$
Perform the multiplication.
$= \boxed{\log 72}$
b) Simplify $2 \log 5$ :
Use the logarithm property $n \log a = \log a^n$ .
$2 \log 5 = \log 5^2$
Calculate the power.
$= \boxed{\log 25}$
c) Simplify $\log 25 \div \log \frac{1}{5}$ :
Rewrite the expression as a fraction and express the arguments as powers of 5.
$\frac{\log 25}{\log \frac{1}{5}} = \frac{\log 5^2}{\log 5^{-1}}$
Use the logarithm property $\log a^n = n \log a$ .
$= \frac{2 \log 5}{-1 \log 5}$
Cancel out $\log 5$ .
$= \boxed{-2}$
2.

a) Simplify $\frac{x^2 - 9}{x^2 + 4x + 15}$ :
Factor the numerator using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ .
$x^2 - 9 = (x-3)(x+3)$
The denominator $x^2 + 4x + 15$ has a discriminant $b^2 - 4ac = 4^2 - 4(1)(15) = 16 - 60 = -44$ .
Since the discriminant is negative, it cannot be factored into linear terms with real coefficients.
The simplified expression is:
$\frac{(x-3)(x+3)}{x^2 + 4x + 15}$
Since no common factors exist, the expression cannot be simplified further.
$= \boxed{\frac{x^2 - 9}{x^2 + 4x + 15}}$
b) Simplify $\frac{2x - 1}{5} - \frac{x - 2}{3}$ :
Find a common denominator, which is $5 \times 3 = 15$ .
$= \frac{3(2x - 1)}{15} - \frac{5(x - 2)}{15}$
Distribute the numbers in the numerators.
$= \frac{6x - 3 - (5x - 10)}{15}$
Distribute the negative sign and combine like terms.
$= \frac{6x - 3 - 5x + 10}{15}$ $= \boxed{\frac{x + 7}{15}}$
3.
A rectangle measures 11 cm by 5 cm correct to the nearest cm.
This means:
Length $L$ : $10.5 \text{ cm} \le L < 11.5 \text{ cm}$
Width $W$ : $4.5 \text{ cm} \le W < 5.5 \text{ cm}$

a) Find the least possible area of the rectangle:
The least possible area occurs when both the length and width are at their least possible values.
Least Area $= L_{min} \times W_{min}$
Least Area $= 10.5 \text{ cm} \times 4.5 \text{ cm}$
Calculate the product.
Least Area $= \boxed{47.25 \text{ cm}^2}$

b) Find the maximum possible perimeter of the rectangle:
The maximum possible perimeter occurs when both the length and width are at their maximum possible values (upper bounds).
Maximum Perimeter $= 2(L_{max} + W_{max})$
Maximum Perimeter $= 2(11.5 \text{ cm} + 5.5 \text{ cm})$
Calculate the sum and then multiply.
Maximum Perimeter $= 2(17 \text{ cm})$
Maximum Perimeter $= \boxed{34 \text{ cm}}$

4.

a) Find the acceleration during the first 10 seconds:
Acceleration is the gradient of the speed-time graph.
For the first 10 seconds, the speed changes from $0 \text{ m/s}$ to $30 \text{ m/s}$ .
Acceleration $= \frac{\text{Change in speed}}{\text{Change in time}}$
Acceleration $= \frac{30 \text{ m/s} - 0 \text{ m/s}}{10 \text{ s} - 0 \text{ s}}$
Calculate the acceleration.
Acceleration $= \frac{30 \text{ m/s}}{10 \text{ s}} = \boxed{3 \text{ m/s}^2}$

b) Total distance travelled:
The total distance travelled is the area under the speed-time graph.
This graph forms a trapezium.
Area of trapezium $= \frac{1}{2} (a+b)h$ , where $a$ and $b$ are the lengths of the parallel sides (time intervals) and $h$ is the height (maximum speed).
Here, $a = (15-10) = 5 \text{ s}$ (top parallel side) and $b = 20 \text{ s}$ (bottom parallel side). The height $h = 30 \text{ m/s}$ .
Total Distance $= \frac{1}{2} (5 \text{ s} + 20 \text{ s}) \times 30 \text{ m/s}$
Calculate the total distance.
Total Distance $= \frac{1}{2} (25 \text{ s}) \times 30 \text{ m/s}$
Total Distance $= 25 \times 15 \text{ m} = \boxed{375 \text{ m}}$

c) Find the speed when time is 15 seconds:
Locate $t=15$ seconds on the time axis of the graph.
Read the corresponding speed value from the vertical axis.
From $t=10$ s to $t=15$ s, the speed is constant.
Speed at $t=15 \text{ s} = \boxed{30 \text{ m/s}}$

5.

a) Find the equation of line P:
Identify two points on line P.
From the graph, line P passes through $(0, 4)$ (y-intercept) and $(6, 0)$ (x-intercept).
Calculate the gradient (slope) $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
$m = \frac{0 - 4}{6 - 0} = \frac{-4}{6} = -\frac{2}{3}$
Use the slope-intercept form $y = mx + c$ , where $c$ is the y-intercept.
Here $c=4$ .
$y = -\frac{2}{3}x + 4$
Convert to general form $Ax + By = C$ (optional, but often preferred).
Multiply by 3 to clear the fraction.
$3y = -2x + 12$ $2x + 3y = \boxed{12}$
b) Find the equation of line Q:
Identify two points on line Q.
From the graph, line Q passes through $(0, -3)$ (y-intercept) and $(6, 0)$ (x-intercept).
Calculate the gradient (slope) $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
$m = \frac{0 - (-3)}{6 - 0} = \frac{3}{6} = \frac{1}{2}$
Use the slope-intercept form $y = mx + c$ , where $c$ is the y-intercept.
Here $c=-3$ .
$y = \frac{1}{2}x - 3$
Convert to general form $Ax + By = C$ .
Multiply by 2 to clear the fraction.
$2y = x - 6$ $x - 2y = \boxed{6}$
6.
A music group has 30 members.
Let F be the set of members who play flute, and M be the set of members who play mbira.
Given:
$n(U) = 30$ (Total members)
$n(F \cap M) = 7$ (Play flute and mbira)
$n(F) = 12$ (Play flute)
$n(F \cup M)' = 1$ (Does not play flute and mbira)

a) Complete the Venn diagram:
Find the number of members who play only flute.
$n(F \text{ only}) = n(F) - n(F \cap M) = 12 - 7 = 5$ .
Find the number of members who play only mbira.
The total number of members is the sum of those who play only flute, only mbira, both, and neither.
$n(U) = n(F \text{ only}) + n(M \text{ only}) + n(F \cap M) + n(F \cup M)'$
$30 = 5 + n(M \text{ only}) + 7 + 1$
$30 = 13 + n(M \text{ only})$
$n(M \text{ only}) = 30 - 13 = 17$ .

The completed Venn diagram values are:
- F only: 5
- F and M: 7
- M only: 17
- Neither: 1
b) Find $n(M)$ (number of members who play mbira):
The number of members who play mbira includes those who play only mbira and those who play both flute and mbira.
$n(M) = n(M \text{ only}) + n(F \cap M)$
Substitute the values found in part (a).
$n(M) = 17 + 7 = \boxed{24}$

7.
Find H.C.F of 36 and 54:
Find the prime factorization of each number.
$36 = 2 \times 18 = 2 \times 2 \times 9 = 2^2 \times 3^2$ $54 = 2 \times 27 = 2 \times 3 \times 9 = 2 \times 3^3$
Identify the common prime factors and their lowest powers.
Common prime factors are 2 and 3.
The lowest power of 2 is $2^1$ .
The lowest power of 3 is $3^2$ .
Multiply these common prime factors raised to their lowest powers.
$\text{H.C.F.} = 2^1 \times 3^2 = 2 \times 9 = \boxed{18}$

Find the H.C.F of 36 and 54

Step-by-step solution

Use the logarithm property log⁡a+log⁡b=log⁡(a×b)\log a + \log b = \log (a \times b)loga+logb=log(a×b).

Perform the multiplication.

Use the logarithm property nlog⁡a=log⁡ann \log a = \log a^nnloga=logan.

Calculate the power.

Rewrite the expression as a fraction and express the arguments as powers of 5.

Use the logarithm property log⁡an=nlog⁡a\log a^n = n \log alogan=nloga.

Cancel out log⁡5\log 5log5.

Factor the numerator using the difference of squares formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b).

The denominator x2+4x+15x^2 + 4x + 15x2+4x+15 has a discriminant b2−4ac=42−4(1)(15)=16−60=−44b^2 - 4ac = 4^2 - 4(1)(15) = 16 - 60 = -44b2−4ac=42−4(1)(15)=16−60=−44.

Find a common denominator, which is 5×3=155 \times 3 = 155×3=15.

Distribute the numbers in the numerators.

Distribute the negative sign and combine like terms.

The least possible area occurs when both the length and width are at their least possible values.

Calculate the product.

The maximum possible perimeter occurs when both the length and width are at their maximum possible values (upper bounds).

Calculate the sum and then multiply.

Acceleration is the gradient of the speed-time graph.

Calculate the acceleration.

The total distance travelled is the area under the speed-time graph.

Calculate the total distance.

Locate t=15t=15t=15 seconds on the time axis of the graph.

Read the corresponding speed value from the vertical axis.

Identify two points on line P.

Calculate the gradient (slope) m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​.

Use the slope-intercept form y=mx+cy = mx + cy=mx+c, where ccc is the y-intercept.

Convert to general form Ax+By=CAx + By = CAx+By=C (optional, but often preferred).

Identify two points on line Q.

Calculate the gradient (slope) m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​.

Use the slope-intercept form y=mx+cy = mx + cy=mx+c, where ccc is the y-intercept.

Convert to general form Ax+By=CAx + By = CAx+By=C.

Find the number of members who play only flute.

Find the number of members who play only mbira.

The number of members who play mbira includes those who play only mbira and those who play both flute and mbira.

Substitute the values found in part (a).

Find the prime factorization of each number.

Identify the common prime factors and their lowest powers.

Multiply these common prime factors raised to their lowest powers.

Still stuck on this one?

More Mathematics Questions

Use the logarithm property $\log a + \log b = \log (a \times b)$ .

Use the logarithm property $n \log a = \log a^n$ .

Use the logarithm property $\log a^n = n \log a$ .

Cancel out $\log 5$ .

Factor the numerator using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ .

The denominator $x^2 + 4x + 15$ has a discriminant $b^2 - 4ac = 4^2 - 4(1)(15) = 16 - 60 = -44$ .

Find a common denominator, which is $5 \times 3 = 15$ .

Locate $t=15$ seconds on the time axis of the graph.

Calculate the gradient (slope) $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

Use the slope-intercept form $y = mx + c$ , where $c$ is the y-intercept.

Convert to general form $Ax + By = C$ (optional, but often preferred).

Calculate the gradient (slope) $m = \frac{y_2 - y_1}{x_2 - x_1}$ .

Use the slope-intercept form $y = mx + c$ , where $c$ is the y-intercept.

Convert to general form $Ax + By = C$ .