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2 years ago

Expressions that simplify to the same expression are called ____ expressions.

Mathematics

2 answers:

larisa86 [58]2 years ago

5 0

They're called equivalent expressions

dusya [7]2 years ago

4 0

Equivalent Expressions are Expressions that simplify to the same expression

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HELP ME PLS W 23 AND 24

n200080 [17]

Answer:

23 is 1/25

24 is -3

Step-by-step explanation:

Please give that brainliest giiirllllll, if you are one thanks tho!!!!!!

add me on snap haha badluckbandit69

8 0

3 years ago

Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vec

Stells [14]

Answer:

$\large\boxed {1.59 mm}$

Explanation:

1. Given vector:

length: 4.00 mm = magnitude of the vector
angle: 23.5º north of east = 23.5º from the x-axys (counterclockwise)

2. y-component

The y-component may be determined using the sine ratio, the angle from the x-axys (counterclockwise direction), and the magnitude of the vector.

sine (23.5º) = y-component / magnitude

y-component = magnitude × sine (23.5º) = 4.00 mm × sine (23.5º) = 1.59 mm.

$\large\boxed{y-component = 1.59 mm}$

7 0

2 years ago

A pet store has 115 fish that need to be placed into fish tanks. Each tank can hold 6 fish. How many tanks does the store need f

galina1969 [7]

Answer:19

Step-by-step explanation: draw the tanks put 6 fish in each tank until you get all 115 in each tank

3 0

2 years ago

Why might someone choose to use the y-intercept and the slope to graph a line?

ss7ja [257]

Because if you know where the line crosses the y axis (the y intercept) you can plot a point there. Then with slope you can count to your next plotted point. With the 2 points you can now draw your line.

4 0

3 years ago

Can anyone figure this out?

Verizon [17]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

5 0

3 years ago