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3 years ago

Find the distance between the two numbers on a number line. -3.1 and 5.7

Mathematics

2 answers:

Assoli18 [71]3 years ago

5 0

Answer:

2.6

Step-by-step explanation:

used a calculator to find out the answer

-3.1 + 5.7 + 2.6

vovikov84 [41]3 years ago

4 0

Answer:

2.6

Step-by-step explanation:

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What is total number of elements in 3 trials of rolling a 6-sided die?

Elina [12.6K]

Answer:

216 elements

Step-by-step explanation:

the total number of elements of a dice is :

6^n

6 stand for the 6 faces of the dice

n stands for the number trial

then total number of elements in 3 trials of rolling a 6-sided die is 6^3=216 elements

3 0

3 years ago

Each cement block weighs 4 3/5 pounds. How much do 3 cement blocks weigh in total?

natali 33 [55]

Answer:

13 4/5 pounds

Step-by-step explanation:

You multiply 4 3/5 by 3, which equals 13.8, which equals 13 4/5 pounds.

3 0

2 years ago

Triangle ABC is translated 4 units down and 6 units right, resulting in triangle A'B'C'.

gogolik [260]

Answer:

The new coordinate of B' will be ( -1, 6)

7 0

3 years ago

Read 2 more answers

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

NO LINKS. Find the measure of the angle or arc indicated. Part 2<br>

sergeinik [125]

9514 1404 393

Answer:

14. C) 136°

15. C) 40°

Step-by-step explanation:

Inscribed angles are half the measure of the arc they intercept. For an inscribed quadrilateral, this means opposite angles are supplementary.

14) ∠H +∠W = 180°

34x +55x +2 = 180

89x = 178 . . . . . . . . . subtract 2

x = 2 . . . . . . . . . . . . . divide by 89

arc VX = 2(34x) = 68(2) = 136 . . . degrees

15) The sum of angles in the triangle is 180°.

? + 80° + (120°/2) = 180°

? = 40° . . . . . . . . . . subtract 140°

6 0

3 years ago

Read 2 more answers