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3 years ago

Multiply the numbers and simplify the answer: 3 x 1- 3/4

1 answer:

5 0

$\underbrace{3\times1}_{first}-\frac{3}{4}=3-\frac{3}{4}=2\frac{1}{4}$

$3\times1\frac{3}{4}=3\times\frac{1\times4+3}{4}=3\times\frac{4+3}{4}=\frac{3}{1}\times\frac{7}{4}=\frac{3\times7}{4}=\frac{21}{4}=\frac{20+1}{4}=\frac{20}{4}+\frac{1}{4}\\\center\boxed{=5\frac{1}{4}}$

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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

What is the truncation error for S4?

solong [7]

Answer:

Its B

Step-by-step explanation:

I calculated it using the formula and got B

3 0

2 years ago

HELP MEEEE PLEASEEEEEEEEEEEEEE

Agata [3.3K]

Answer:

a = 4

Step-by-step explanation:

a^2 + b^2 = c^2

a= ?

b= 3

c= 5 (c is always the hypotenuse)

*plug in given values

a^2 + 3^2 = 5^2

a^2 + 9 = 25

-9 -9

a^2 = 16

*find the square root

sqrt(a) = sqrt(16)

a = 4

8 0

2 years ago

Read 2 more answers

Mutiply 4y-4w^8-3y^5w^3

sukhopar [10]

Answer:

48y⁶w¹¹

Step-by-step explanation:

[-3y⁵][4y] → -12y⁶

[w³][-4w⁸] → -4w¹¹

______

48y⁶w¹¹

I am joyous to assist you anytime.

8 0

3 years ago

-6×f (-3) -5×g(-7) =

polet [3.4K]

Answer:

=18fx+35gx

Step-by-step explanation:

I think.. Ask your teacher if it's right have a good day :)

4 0

3 years ago