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3 years ago

8

A student said we cannot subtract 1.97 from 20 because one 1.97 has two decimal digits and 20 has none do you agree with this st

atement Explain or show your reasoning

1 answer:

Pie3 years ago

5 0

Answer: We can

Step-by-step explanation: 20.00

- 1.97

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Can someone please help me with this

oee [108]

Answer:

165 pennies

Step-by-step explanation:

250/1.52=164.473684211, but just round it to the nearest whole number whitch would be 165!

6 0

2 years ago

Help asap will mark brainliest

katen-ka-za [31]

Answer:

(-5 3/4, 1 2/3)

Step-by-step explanation:

the line intersects right before the -6, so it’s not -6. yet, it is as close to -6 as you can get which would be -5 3/4. meaning the closes would be the third choice.

8 0

2 years ago

Solve the following problems using 5-D process part 2

svet-max [94.6K]

The two problems are almost identical: you have to set up a system, and then solve it.

In the first case, let $g$ and $b$ be, respectively, the number of girls and boys.

We know that $g=2b+4$ (the number of girls in the Spanish Club is four more than twice the number of boys). Also, we know that $g+b=61$ (there are 61 students in total).

So, we have the system

$\begin{cases}g=2b+4\\g+b=61\end{cases}$

We can use the first equation to substitute in the second

$g+b=61 \iff (2b+4)+b=61 \iff 3b+4=61 \iff 3b=57 \iff b=19$

And then solve for $g$ :

$g=2b+4=2\cdot 19+4=38+4=42$

For the second problem, let $c$ and $j$ be the number of pins knocked down by, respectively, Carrie and John. Just like before, we have the system

$\begin{cases}c=j+12\\c+j=230\end{cases}$

And you can solve it in the very same way we solved the previous one.

5 0

3 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago

PLEASE HELP!!! 15 POINTS!!

tankabanditka [31]

Answer:

23.4 cm

Step-by-step explanation:

If the base of the triangle is half of that of a given side, then we need to divide the given side by 2 to get the base.

46.8

---------2

23.4

There is your answer

Hope it helped

Bye

7 0

2 years ago

Read 2 more answers