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3 years ago

Round each number to the nearest ten thousand and then subtracting

Mathematics

1 answer:

katrin [286]3 years ago

7 0

Answer:

43,000

Step-by-step explanation:

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find the measure of the arc or angle indicated. assume that lines which appear to be diameters are actual diameters just enter t

konstantin123 [22]

Answer:

I think D

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3 years ago

Two parallel lines are crossed by a transversal

Olegator [25]

B. H=60
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3 years ago

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0

3 years ago

What is the equation of this line? y=23x y=32x y=−32x y=−23x

expeople1 [14]

y= 2/3x

Please consider marking brainliest! :)

4 0

3 years ago

Practice working with reflections.

Alinara [238K]

Answer: The rule of the reflection is rx-axis(x, y) → (x, –y) ⇒ 3rd answer

Step-by-step explanation:

Let us revise the reflection on the axes

If point (x , y) reflected across the x-axis , then its image is (x , -y) , the rule of reflection is rx-axis (x , y) → (x , -y)

If point (x , y) reflected across the y-axis , then its image is (-x , y) , the rule of reflection is ry-axis (x , y) → (-x , y)

In Δ LMN

∵ L = (-4 , 2)

∵ M = (-5 , 4)

∵ N = (-2 , 3)

In ΔL'M'N'

∵ L' = (-4 , -2)

∵ M' = (-5 , -4)

∵ N' = (-2 , -3)

The signs of y-coordinates of the vertices of Δ LMN are changed, that means Δ LMN are reflected across the x-axis

∵ The y-coordinate of L is 2 and the y-coordinate of L' is -2

∵ The y-coordinate of M is 4 and the y-coordinate of M' is -4

∵ The y-coordinate of N is 3 and the y-coordinate of N' is -3

∴ Δ LMN is reflected across the x-axis

∴ The image of point (x , y) is (x , -y)

3 0

3 years ago

Read 2 more answers