All categories

3 years ago

Just # 16idk what it's "counting" by

Mathematics

1 answer:

zysi [14]3 years ago

3 0

For #16, every input by +1 has an output of +7.

9 + 7 = 16
16 + 7 = 23
23 + 7 = 30.....

You might be interested in

If set A has 4 elements and B has 3 elements then set n(AXB) is a) 12 b) 14 c) 24 d) 7

Karo-lina-s [1.5K]

Answer:

The correct option is a.

Step-by-step explanation:

Given information:

Number of elements in set A = 4

Number of elements in set B = 3

We need to find n(A × B). It means we have to find the number of elements in set A × B.

If P and Q are two sets with elements m and n respectively, then the number of element in the set P × Q is m × n.

$n(A\times B)=n(A)\times n(B)$

$n(A\times B)=4\times 3$

$n(A\times B)=12$

The value of n(A × B) is 12. Therefore the correct option is a.

8 0

3 years ago

London mixes 6 parts yellow paint with 3 parts blue paint to make green. She adds 1 more part of yelow to change the

Stella [2.4K]

Answer:

The new ratio is $\frac{7}{3}$ ⇒ (D)

Step-by-step explanation:

<em>The given information is</em>

6 parts of yellow paint with 3 parts of blue paints to make green paint
The yellow paint adds by 1 part

We need to find the ratio of yellow to blue

<em>Let us find the correct answer</em>

∵ The yellow = 6 parts

∵ The yellow adds by 1 part

∴ The yellow = 6 parts + 1 parts = 7 parts

∵ The blue = 3 parts

∵ The ratio of yellow to blue = $\frac{yellow}{blue}$

∴ The ratio of yellow to blue = $\frac{7}{3}$

The new ratio is $\frac{7}{3}$

7 0

3 years ago

A rectangle has a perimeter of 26 cm and one of its sides has length of 5 cm

otez555 [7]

Answer:5 by 8 by 5 by 8

Step-by-step explanation: the perimeter is 26 and one side if 5 so you know that another side is also 5. So add 5 and 5 together you get 10. Then subtract 10 from the perimeter(witch we know is 26) and you end up with 16. we know that the last two side lengths are the same so divide 16 by two and the sidle length is 8.

8 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

Covering the brake can cut Reaction Time by what percent

Allisa [31]

The percebt decides what rate of change it js so okease give examples

6 0

3 years ago