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Artyom0805 [142]

3 years ago

13

if you are the first one to answer this question, you will be marked brsinlyest (ik I spelled that wrong but yk what I mean) wha

t is 20+20x17-12 use mdas (my dear aunt sally) to answer.

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

348

Step-by-step explanation:

Multiply 20 x 17 first. Then add that value to 20 and -17. The result is the answer.

I get 348

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It took emily 25 min to rideHer bicycle to the repair shop in one hour and 15 minutes to walk back home. If Emily her bicycle to

s2008m [1.1K]

Answer:the repair shop is 5km from her house.

Step-by-step explanation:

Let x represent the distance of the repair shop from her house.

Let y represent Emily's speed when riding.

Distance = speed × time

It took Emily 25 min to ride her bike to the repair shop. Converting 25 min to hours, it becomes 25/60 = 5/12 hours.

Distance covered,

x = y × 5/12 = 5y/12

It took her 1h 15 min to walk back home. Converting to hours, it becomes 1 + 15/60 = 1 1/4 = 5/4

If she can ride her bike 8km/h faster than she can walk, it means that her speed while walking would be y 8 8

Therefore,

Distance covered,

x= 5/4(y - 8) = (5y - 40)/4

Since the distance remains the same, then

5y/12 =(5y - 40)/4

Crossmultiplying, it becomes

5y × 4 = 12(5y - 40)

20y = 60y - 480

60y - 20y = 480

40y = 480

y = 480/40 = 12

x = 5y/12 = 5 × 12/12 = 5 km

6 0

2 years ago

What does 6(-2.1k - 2) + (7k + 5) equal

ikadub [295]

Answer:

Step-by-step explanation:

6(-2.1k-2)+(7k+5)

-12.6k-12+7k+5

-5.6k-7

5 0

3 years ago

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:

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

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3 years ago

Please help I’m failing

vredina [299]

The answer is B.........

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What is the pattern in the values as the exponents increase?

likoan [24]

Answer:

The answer is D. Multiply the previous value by 3.

Step-by-step explanation:

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2 years ago

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