Question

Edit the following sentences to eliminate problems with subject-verb agreement and write the edited sentence.
If a sentence is correct, write “correct.”

Jack’s first days in the infantry was grueling.
Incorrect
-Jack’s first days in the infantry were grueling.

1. One of the main reasons for elephant poaching are the profits received from selling the ivory tusks.
one of the main reason for elephant's poaching is the profit received by selling ivory tusks.
2. Not until my interview with Dr. Chang were other possibilities opened to me.
Until my interview with Dr. Chang other opportunities were not opened to me.
3. Batik cloth from Bali, blue and white ceramics from Cambodia, and a bocce ball from Turin has made Hannah’s room the talk of the dorm.


4. The board of directors, ignoring the wishes of the neighbourhood, has voted to allow further development.
_________
5. Measles is a contagious childhood disease.
Measles are contageous childhood disease.
6. The presence of certain bacteria in our bodies are one of the factors that determines our overall health.
_________
7. Leah is the only one of the many applicants who has the ability to step into this job.
_________
8. Neither the explorer nor his companions was ever seen again.
_________

Answer 1

Answer:

$\large\underline {\sf {\pink{ \pink\star {\: GIVEN: - }}}}$

$\sf{ \rightarrow \: If \: a \: and \: B are \: the \: roots \: of \: the \: equation \: } \\ \sf{ ax ^{2} + bx + c =0,}</p><p>$

$\large\underline {\sf {\red{ \pink\star {\: TO \: FIND : - }}}}$

$\sf{ \rightarrow \: express \: the \: roots \: of \: the \: equation \:} \\ \sf{ {a}^{3} {x}^{2} - ab ^{2} x + b ^{2} c = 0 \: in \: terms \: of \: } \\ \sf{a \: and \: B \: .}</p><p>$

$\large\underline {\sf {\orange{ \pink\star {\: SOLUTION \: : - }}}}$

$\sf\green{ \alpha \: and \beta \: are \: the \: roots \: of \: the \: equation. \: }$

$\sf{ \Rightarrow \: ax ^{2} + bx + c = 0.}$

$\sf \blue{As \: we \: know \: that,}$

$\sf\green{Sum \: of \: the \: zeroes \: of \: the \: quadratic } \\ \sf{ \: polynomial.} \\ </p><p> \sf{ \Rightarrow\alpha + \beta = \frac{ - b}{a} }$

$\sf\green{Products \: of \: the \: zeroes \: of \: the \: quadratic \: } \\\sf\green{ polynomial.} \\ \sf{ \Rightarrow \: \alpha \beta = \frac{c}{a} }$

$\sf\green{Roots \: of \: the \: equation}$

$\sf{ \Rightarrow \: a ^{3} \: x ^{2} ab ^{2} x + b^{2} c = 0.}$

$\sf \blue{As \: we \: know \: that,}$

$\sf\green{Let, \:y \:and\: \delta \:are \:the \:roots\: of \:the \:equation.}$

$\sf{\Rightarrow a ^{3} \: x ^{2} - ab ^{2} x + b^{2}c = 0.}$

$\sf\green{Sum \:of\: the \:zeroes \:of \:the \:quadratic\: }$

$\sf\green{polynomial.}$

$\sf{\Rightarrow \:y + \delta = \frac{-b}{a}}$

$\sf{\Rightarrow \:y + \delta=\frac{(-ab^{2})}{a^{3}} = ab^{2}/a^{3}.}$

$\sf{\Rightarrow \:y +\delta = =(\: \frac{b^{2}}{a^{2}}\: )=(\: \frac{-b}{a}^{2}\: )= (\: \alpha +\beta \: )^{2}.}$

$\sf\green{Products \:of\: the \:zeroes \:of \:the \:quadratic}$

$\sf\green{polynomial.}$

$\sf{\Rightarrow \: y \delta = \frac{c}{a}}$

$\sf{\Rightarrow \: y \delta =(\frac{b^{2}c}{a^{3}})=(\frac{b^{2}}{a^{2}}) (\frac{c}{a}) }$

$\sf{\Rightarrow \:y \delta =(\: \frac{b}{a}^{2}\: )(\frac{c}{a})= (\: \alpha +\beta \: )^{2}(\: \alpha +\beta \: ) }$