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3 years ago

What is the location of the point on the number line that is 2/7 of the way from A=18 to B=4?

Mathematics

2 answers:

olchik [2.2K]3 years ago

5 0

Answer:

I believe that it would be 14

Step-by-step explanation:

Over [174]3 years ago

5 0

Answer:

14 is the right answer

Step-by-step explanation:

Given that on the number line A=14 and B =4

The movement starts from A to B and travels 2/7 of distance of AB

AB = distance between A and B = 14

2/7 of AB = 2/7 (14) = 4

Hence when starting from A travel towards number line a distance of 4 units means the point would be 4 to the left of A, i.e 14

So new position would be 14.

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Malik's bike tire has a circumference of 25.12 inches. What is the radius of the bike tire?

Elan Coil [88]

4 because you divide it first and then you divided by 3.14

6 0

3 years ago

Express the following equation in slope intercept form: 8x-2y=10 pleaseeeeeeeeeeee

Sergeu [11.5K]

Answer:

y = 4x–5

Step-by-step explanation:

Slope intercept form is y = mx +b. We can rearrange the equation to fit slope intercept form.

–2y = –8x + 10 (we get this by subtracting 8x from both sides)

In order to isolate the y, we divide both sides by –2.

$\frac{-2y}{-2}$ = $\frac{-8x}{-2}$ + $\frac{-10}{-2}$

–2 divided by –2 equals 1, –8 divided by –2 equals 4, and 10 divided by –2 equals 5.

Therefore, the answer is y = 4x–5.

8 0

3 years ago

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations. Hint: Use

QveST [7]

$\bf 27\implies 3^3\\\\
i^3\implies i\cdot i\cdot i\implies i^2\cdot i\implies -1\cdot i\implies -i\\\\
-----------------------------\\\\
x^3-27i=0\implies x^3+3^3(-i)=0\implies x^3+(3^3i^3)=0
\\\\\\
x^3+(3i)^3=0\\\\
-----------------------------\\\\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3\\\\
-----------------------------\\\\$

$\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies 
\begin{cases}
x+3i=0\implies \boxed{x=-3i}\\\\
x^2-3ix+(3i)^2=0
\end{cases}
\\\\\\
\textit{now, for the second one, we'd need the quadratic formula}
\\\\\\
x^2-3ix+(3i)^2=0\implies x^2-3ix+(3^2i^2)=0
$

$\bf x^2-3ix+(-9)=0\implies x^2-3ix-9=0
\\\\\\
\textit{quadratic formula}\\\\
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\implies x=\cfrac{3i\pm\sqrt{(-3i)^2-4(1)(-9)}}{2(1)}
\\\\\\
x=\cfrac{3i\pm\sqrt{(-3)^2i^2+36}}{2}\implies x=\cfrac{3i\pm\sqrt{-9+36}}{2}
\\\\\\
x=\cfrac{3i\pm\sqrt{27}}{2}\implies \boxed{x=\cfrac{3i\pm 3\sqrt{3}}{2}}$

7 0

3 years ago

What are the zeros for both functions

sammy [17]

Answer:

the zero of a function is the value of x which makes the final value zero

First Equation:

So let 25 - 2x equal to 0

25 - 2x = 0

x = 12.5

Second Equation:

Let 2x² - 11x - 6 equal to zero

2x² - 11x - 6 = 0

2x² - 12x + x - 6 = 0 (Splitting the middle term)

2x(x - 6) + 1(x - 6) = 0

(2x + 1) (x - 6) = 0

So we can transpose either one of the brackets below the zero

(2x + 1) = 0 or (x-6) = 0

x = -1/2 or x = 6

5 0

4 years ago

Use the graphing calculator to locate the solutions of this system of equations: y = x2 + x – 2 y = –x + 1 What are the solution

pishuonlain [190]

The solutions are (-3,4) and (1,0)

3 0

3 years ago