The product of two integers is -160 if one of them is 8 find the other

Answers

Answer 1

Hey ! there

Answer:

Other number is -20

Step-by-step explanation:

In this question we are given that the product of two integers is -160 and one of them is 8 . And we are asked to find the other integer .

For finding the other integer we are assuming ,

Other integer as ' x '

And now we can find other Integer by making equation and solving it .

SOLUTION : -

Equation is :

[tex] \quad \dashrightarrow \qquad \: 8 \times x = - 160[/tex]

or ,

[tex] \quad \dashrightarrow \qquad \: 8x = - 160[/tex]

To isolate the variable x , We are dividing both sides by 8 :

[tex] \quad \dashrightarrow \qquad \: \dfrac{ \cancel{8}x}{ \cancel{8}} = \cancel{ \dfrac{ - 160}{8} }[/tex]

On simplifying it , We get :

[tex] \quad \dashrightarrow \qquad \: \red{\underline{\boxed{\frak{x = - 20}}}} \quad \bigstar[/tex]

Henceforth , the other integer is ❝ -20 ❞.

Verifying : -

Now we are verifying whether our answer is wrong or right by substituting value of x in the equation . So ,

[tex]8 \times x = - 160[/tex]

Substituting value of x :

[tex]8 \times - 20 = - 160[/tex]

Multiplying 8 with - 20 :

[tex] - 160 = - 160[/tex]

[tex] \rm{L.H.S = R.H.S}[/tex]

[tex] \rm{Hence , Verified .}[/tex]

Therefore , of answer is correct .

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Answer 2

Answer:

The other integer is -20

[tex] \: [/tex]

Step-by-step explanation:

[tex] \: [/tex]

Given,

The product of two integers is -160.

One of the integers is 8

[tex] \: [/tex]

To Find,

The other integer.

[tex] \: [/tex]

Solution,

Let us assume the other integer as x

[tex] \: [/tex]

So, The equation will be :

[tex] \\ \longrightarrow \qquad{ \ {{ { \sf{x \times 8= - 160}}}}} \\ \: \: [/tex]

[tex] \longrightarrow \qquad{ \ {{ { \sf{8x= - 160}}}}} \\ \: \: [/tex]

Now, Let us divide both sides by 8 :

[tex] \\ \longrightarrow \qquad{ \ {{ { \sf{ \frac{8x}{8} = \frac{ - 160}{8} }}}}} \\ \: \: [/tex]

[tex] \longrightarrow \qquad{ \underline {\boxed{ \pmb{ \mathfrak{x= - 20}}}}} \: \: \bigstar \: \\ [/tex]

Therefore,

The other integer is -20

Related Questions

PLS HELP!! I THIS IS WORTH 200 PTS!! I WILL MARK BRAINIEST WHEN SOMEONE ANSWERS FIRST

Answers

14)

Expression: 5z² - 2z + 1 = 0

Discriminant formula: b²- 4ac

Identify: a = 5, b = -2, c = 1

Solve for discriminant: (-2)²- 4(5)(1) = 4 - 20 = -16

15)1st equation: 1m - 1p = 2

p = 1 also can be rewritten as 2nd equation: 0m + 1p = 1

Build Augment Matrices by taking coefficient's:

[tex]\left[\begin{array}{ccc}1&-1&|| \ \ 2\\0&1&|| \ \ 1\end{array}\right][/tex]

Solve:

Add row 2 to row 1 : R₁ = R₁ + R₂ and restore it back

[tex]\rightarrow \ \left[\begin{array}{ccc}1+0&-1+1& \ \ \ \ || \ 2+1\\0&1&|| \ \ 1\end{array}\right][/tex]

[tex]\rightarrow \ \left[\begin{array}{ccc}1&0& || \ \ 3\\0&1&|| \ \ 1\end{array}\right] \ \ \ = The \quad reduced \quad matrix[/tex]

Solution = (m, p) = (3, 1)

Answer:

14. -16

15. m = 3 and p = 1

Step-by-step explanation:

Question 14

Discriminant

[tex]b^2-4ac\quad\textsf{when}\quad ax^2+bx+c=0[/tex]

Given equation: [tex]5z^2-2z+1=0[/tex]

Therefore:

a = 5b = -2c = 1

Substituting the values into the discriminant:

[tex]\begin{aligned}\implies b^2-4ac & = (-2)^2-4(5)(1)\\& = 4-20\\& = -16\end{aligned}[/tex]

As the discriminant is negative, this implies there are no real roots.

Question 15

Augmented Matrices

[tex]\begin{array}{c}ax+by=p \\cx+dy=q\end{array} \implies \left[\begin{array}{r r | r}a & b & p \\ c & d & q\end{array}\right][/tex]

Use elementary row operations to convert into:

[tex]\left[\begin{array}{r r | r}1 & 0 & h \\ 0 & 1 & k\end{array}\right] \implies \begin{array}{r}x=h\\y=k\end{array}[/tex]

Given:

[tex]\begin{aligned}m-p & =2\\ p & = 1\end{aligned} \implies \left[\begin{array}{r r | r}1 & -1 & 2 \\ 0 & 1 & 1\end{array}\right][/tex]

[tex]\left[\begin{array}{r r | r}1 & -1 & 2 \\ 0 & 1 & 1\end{array}\right] \begin{array}{c}R_1 + R_2 \rightarrow R_1 \\\rightarrow\end{array} \left[\begin{array}{r r | r}1 & 0 & 3 \\ 0 & 1 & 1\end{array}\right][/tex]

Therefore:

[tex]m=3[/tex]

[tex]p=1[/tex]

In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC 45°, Find ∠BCD.