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3 years ago

Are the terms 2s and 3s² like terms?

Mathematics

1 answer:

adelina 88 [10]3 years ago

6 0

Answer:

No.

Step-by-step explanation:

2s is different from 3s squared. Because 2s is not squared.

I hope this helped you.

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Can someone please help with this it’s worth 19 points?

nexus9112 [7]

Answer:

54, 36

Step-by-step explanation:

67+59+x=60*3

126+x=180

x=54

54 cones on Sunday

_____________________________________

180+x=54*4

180+x=216

x=36

36 cones on Monday

6 0

2 years ago

What is 4/9 closest to 0,1/2 or 1

topjm [15]

Answer:

1/2

Step-by-step explanation:

4/9 is equal to 4.444 repeating so that would be closest to 1/2

4 0

2 years ago

Read 2 more answers

HELP A line passes through two points with coordinates (6,-8) and (-3,7).

FromTheMoon [43]

Answer:

3y+5x=6

Step-by-step explanation:

Equation of the Line

The equation of a line passing through points (x1,y1) and (x2,y2) can be found as follows:

$\displaystyle y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The line passes through the points (6,-8) and (-3,7), thus:

$\displaystyle y+8=\frac{7+8}{-3-6}(x-6)$

$\displaystyle y+8=\frac{15}{-9}(x-6)$

Simplifying:

$\displaystyle y+8=-\frac{5}{3}(x-6)$

Multiplying by 3:

$3(y+8)=-5(x-6)$

$3y+24=-5x+30$

Moving all the variables to the left side:

3y + 5x = 30 - 24

3y + 5x = 6

4 0

2 years ago

What is the probability that a dart will hit a spot in the shaded circle?

amid [387]

Answer:

I would say if you at least try to make it then you will end up making it

Step-by-step explanation:

3 0

2 years ago

D Evaluate arcsin (6)] at x = 4. dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

2 years ago