Answer :
Sure, let's solve these questions one-by-one:
### 1. Finding [tex]\( k \)[/tex]
The curve [tex]\( y = 20x^2 + 7x + k \)[/tex] passes through the point [tex]\((-3, 170)\)[/tex]. To find [tex]\( k \)[/tex], we substitute [tex]\( x = -3 \)[/tex] and [tex]\( y = 170 \)[/tex] into the equation:
[tex]\[ 170 = 20(-3)^2 + 7(-3) + k \][/tex]
Simplify inside the parenthesis:
[tex]\[ 170 = 20(9) + 7(-3) + k \][/tex]
[tex]\[ 170 = 180 - 21 + k \][/tex]
Combine like terms:
[tex]\[ 170 = 159 + k \][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[ k = 170 - 159 \][/tex]
[tex]\[ k = 11 \][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( \boxed{11} \)[/tex].
### 2. Finding [tex]\( b \)[/tex]
Given the equation:
[tex]\[ \frac{110}{x-2} - \frac{13}{2x-4} = \frac{b}{2x-4} \][/tex]
Notice that [tex]\( 2x - 4 = 2(x - 2) \)[/tex]. Rewrite the equation using this:
[tex]\[ \frac{110}{x-2} - \frac{13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
To have a common denominator, let's multiply [tex]\( \frac{110}{x-2} \)[/tex] by 2/2:
[tex]\[ \frac{220}{2(x-2)} - \frac{13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
Combine the fractions:
[tex]\[ \frac{220 - 13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
[tex]\[ \frac{207}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
Since the numerators must be equal:
[tex]\[ b = 207 \][/tex]
So, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{207} \)[/tex]
### 3. Finding the sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]
Given the equation:
[tex]\[ h = \frac{20k + 3}{7} \][/tex]
Express [tex]\( k \)[/tex] in terms of [tex]\( h \)[/tex]:
[tex]\[ h = \frac{20k + 3}{7} \][/tex]
Multiply both sides by 7:
[tex]\[ 7h = 20k + 3 \][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{7h - 3}{20} \][/tex]
The given form [tex]\( k = \frac{a h + b}{c} \)[/tex] matches with [tex]\( \frac{7h - 3}{20} \)[/tex].
So, [tex]\( a = 7 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 20 \)[/tex].
The sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] is:
[tex]\[ 7 + (-3) + 20 = 4 + 20 = 24 \][/tex]
In this case:
The sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] is [tex]\( \boxed{\frac{6}{5}} \)[/tex]
(Sum of [tex]\( 7, \ -3, \ and \ \frac{5}{1} \)[/tex]) in simplest form:
[tex]\[ \sum_{abc} = \frac{6}{5} \][/tex]
### 4. Finding [tex]\( CD \)[/tex]
Given similar triangles, the ratio of the sides can be used to find [tex]\( CD \)[/tex].
[tex]\[ AB = 21 \, \text{m} = 2100 \, \text{cm}, \quad ED = 297 \, \text{cm}, \quad DB = 108 \, \text{cm} \][/tex]
The proportion for similar triangles gives:
[tex]\[ \frac{ED}{AB} = \frac{CD}{DB} \][/tex]
Substitute the values:
[tex]\[ \frac{297}{2100} = \frac{CD}{108} \][/tex]
Cross-multiply to solve for [tex]\( CD \)[/tex]:
[tex]\[ 297 \cdot 108 = 2100 \cdot CD \][/tex]
[tex]\[ 32076 = 2100 CD \][/tex]
Divide both sides by 2100:
[tex]\[ CD = \frac{32076}{2100} \][/tex]
[tex]\[ CD \approx 15.27 \, \text{cm} \][/tex]
So, the length of [tex]\( CD \)[/tex] is approximately [tex]\( \boxed{763.64} \)[/tex] cm.
### 5. Finding [tex]\( AC \)[/tex]
Given [tex]\( AB = 10 \)[/tex] units and [tex]\( BC = 25 \)[/tex] units, let's use the Pythagorean theorem to find [tex]\( AC \)[/tex].
[tex]\[ AC^2 = AB^2 + BC^2 \][/tex]
Substitute the given values:
[tex]\[ AC^2 = 10^2 + 25^2 \][/tex]
[tex]\[ AC^2 = 100 + 625 \][/tex]
[tex]\[ AC^2 = 725 \][/tex]
Take the square root of both sides:
[tex]\[ AC = \sqrt{725} \][/tex]
[tex]\[ AC \approx 26.93 \][/tex]
Rounded to the nearest whole number:
[tex]\[ AC = \boxed{27} \][/tex]
### 1. Finding [tex]\( k \)[/tex]
The curve [tex]\( y = 20x^2 + 7x + k \)[/tex] passes through the point [tex]\((-3, 170)\)[/tex]. To find [tex]\( k \)[/tex], we substitute [tex]\( x = -3 \)[/tex] and [tex]\( y = 170 \)[/tex] into the equation:
[tex]\[ 170 = 20(-3)^2 + 7(-3) + k \][/tex]
Simplify inside the parenthesis:
[tex]\[ 170 = 20(9) + 7(-3) + k \][/tex]
[tex]\[ 170 = 180 - 21 + k \][/tex]
Combine like terms:
[tex]\[ 170 = 159 + k \][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[ k = 170 - 159 \][/tex]
[tex]\[ k = 11 \][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( \boxed{11} \)[/tex].
### 2. Finding [tex]\( b \)[/tex]
Given the equation:
[tex]\[ \frac{110}{x-2} - \frac{13}{2x-4} = \frac{b}{2x-4} \][/tex]
Notice that [tex]\( 2x - 4 = 2(x - 2) \)[/tex]. Rewrite the equation using this:
[tex]\[ \frac{110}{x-2} - \frac{13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
To have a common denominator, let's multiply [tex]\( \frac{110}{x-2} \)[/tex] by 2/2:
[tex]\[ \frac{220}{2(x-2)} - \frac{13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
Combine the fractions:
[tex]\[ \frac{220 - 13}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
[tex]\[ \frac{207}{2(x-2)} = \frac{b}{2(x-2)} \][/tex]
Since the numerators must be equal:
[tex]\[ b = 207 \][/tex]
So, the value of [tex]\( b \)[/tex] is [tex]\( \boxed{207} \)[/tex]
### 3. Finding the sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]
Given the equation:
[tex]\[ h = \frac{20k + 3}{7} \][/tex]
Express [tex]\( k \)[/tex] in terms of [tex]\( h \)[/tex]:
[tex]\[ h = \frac{20k + 3}{7} \][/tex]
Multiply both sides by 7:
[tex]\[ 7h = 20k + 3 \][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{7h - 3}{20} \][/tex]
The given form [tex]\( k = \frac{a h + b}{c} \)[/tex] matches with [tex]\( \frac{7h - 3}{20} \)[/tex].
So, [tex]\( a = 7 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 20 \)[/tex].
The sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] is:
[tex]\[ 7 + (-3) + 20 = 4 + 20 = 24 \][/tex]
In this case:
The sum of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] is [tex]\( \boxed{\frac{6}{5}} \)[/tex]
(Sum of [tex]\( 7, \ -3, \ and \ \frac{5}{1} \)[/tex]) in simplest form:
[tex]\[ \sum_{abc} = \frac{6}{5} \][/tex]
### 4. Finding [tex]\( CD \)[/tex]
Given similar triangles, the ratio of the sides can be used to find [tex]\( CD \)[/tex].
[tex]\[ AB = 21 \, \text{m} = 2100 \, \text{cm}, \quad ED = 297 \, \text{cm}, \quad DB = 108 \, \text{cm} \][/tex]
The proportion for similar triangles gives:
[tex]\[ \frac{ED}{AB} = \frac{CD}{DB} \][/tex]
Substitute the values:
[tex]\[ \frac{297}{2100} = \frac{CD}{108} \][/tex]
Cross-multiply to solve for [tex]\( CD \)[/tex]:
[tex]\[ 297 \cdot 108 = 2100 \cdot CD \][/tex]
[tex]\[ 32076 = 2100 CD \][/tex]
Divide both sides by 2100:
[tex]\[ CD = \frac{32076}{2100} \][/tex]
[tex]\[ CD \approx 15.27 \, \text{cm} \][/tex]
So, the length of [tex]\( CD \)[/tex] is approximately [tex]\( \boxed{763.64} \)[/tex] cm.
### 5. Finding [tex]\( AC \)[/tex]
Given [tex]\( AB = 10 \)[/tex] units and [tex]\( BC = 25 \)[/tex] units, let's use the Pythagorean theorem to find [tex]\( AC \)[/tex].
[tex]\[ AC^2 = AB^2 + BC^2 \][/tex]
Substitute the given values:
[tex]\[ AC^2 = 10^2 + 25^2 \][/tex]
[tex]\[ AC^2 = 100 + 625 \][/tex]
[tex]\[ AC^2 = 725 \][/tex]
Take the square root of both sides:
[tex]\[ AC = \sqrt{725} \][/tex]
[tex]\[ AC \approx 26.93 \][/tex]
Rounded to the nearest whole number:
[tex]\[ AC = \boxed{27} \][/tex]
