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2 years ago

A class writes the equation n + n + 1 + n + 2 = 87 to solve the following problem.

Mathematics

1 answer:

Bess [88]2 years ago

8 0

Answer:

n + n+2 + m+4 = 87

I got this answer.....hope it may helpful

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In triangle ABC, angle A is 25 degree, and Angle B is 105 degrees. what is the measure of angle c

Scilla [17]

Angle c= 60°

This is because 105+25+x=180°
Hope this helps and good luck!! :)

8 0

2 years ago

My product is 30. The differencd of the two factors is 1. The zum of the two factors is 11. What numbers am i

timurjin [86]

A * b = 30
a - b = 1
a + b = 11

take ur last 2 equations, and add them
a - b = 1
a + b = 11
--------------add
2a = 12
a = 12/2
a = 6

now its just a matter of subbing
a + b = 11
6 + b = 11
b = 11 - 6
b = 5

so a = 6 and b = 5...whose product is 30, whose difference is 1, and whose sum is 11.

5 0

2 years ago

Find all the zeros for each function P(x)=x^4-4x^3-x^2+20x-20

ivanzaharov [21]

Answer:

The zeros of the given polynomial function are

2,2, $\pm\sqrt{5}$

Step-by-step explanation:

Given polynomial is $P(x)=x^4-4x^3-x^2+20x-20$

To find the zeros equate the given polynomial to zero

ie., P(x)=0

$P(x)=x^4-4x^3-x^2+20x-20=0$

By using synthetic division we can solve the polynomial:

2_| 1 -4 -1 20 -20

0 2 -4 -10 20

_____________________

1 -2 -5 10 |_0

Therefore x-2=0

x=2 is a zero of P(x)

Now we can write the cubic equation as below:

$x^3-2x^2-5x+10=0$

Again using synthetic division

2_| 1 -2 -5 10

0 2 0 -10

______________

1 0 -5 |_0

Therefore x-2=0

x=2 is also a zero of P(x).

Now we have $x^2+0x-5=0$

$x^2-5=0$

$x^2=5$

$x^=\pm\sqrt{5}$ is a zero of P(x)

Therefore the zeros are 2,2, $\pm\sqrt{5}$

8 0

3 years ago

A line passes through the points (2,4) and (5,6) . Select Yes or No to tell whether each equation describes this line. Equation

krek1111 [17]

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-4}{5-2}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=\cfrac{2}{3}(x-2)$

3 0

3 years ago

Which sequence of similar transformations could map △ABC onto △A'B'C'?

nadya68 [22]

Answer: A sequence of similar transformations of dilation and translation could map △ABC onto △A'B'C'.

Step-by-step explanation:

Similar transformations: If one figure can be mapped onto the other figure using a dilation and a congruent rigid transformation or a rigid transformation followed by dilation then the two figures are said to be similar.

In the attachment △ABC mapped onto △A'B'C' by a sequence of dilation from origin and scalar factor k followed by translation.

7 0

3 years ago

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