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3 years ago

Find two integers whose sum is -13 and product is 40

Mathematics

2 answers:

Nastasia [14]3 years ago

7 0

-8 and -5

-8 x -5 = 40
-8 + (-5) = -13

Umnica [9.8K]3 years ago

3 0

Let us say, the numbers are "a" and "b"
their sum is -13
a + b = -13

their product is 40
a * b = 40 $\bf \begin{cases}
a+b=-13\\
a\cdot b=40\implies b=\frac{40}{a}\\
then\\
------------\\
a+\boxed{b}=-13\implies a+\boxed{\frac{40}{a}}=-13
\end{cases}$

solve for "a",
what is "b"? well, b = 40/a :)

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2) If 3/5 of students carry an iPhone, what fraction of students is this using 75 as the denominator

kkurt [141]

Answer:

45/75

Step-by-step explanation:

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8 0

3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago

Can someone please help me

fgiga [73]

JKL should be the right answer

5 0

2 years ago

PLEASE HELP ASAPPPP!

Gemiola [76]

Answer:

3 7/16 cups

Step-by-step explanation: