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4 years ago

Two equivalent decimals for 2.50

1 answer:

5 0

2.5 is an equivalent decimal because you can add infinity zeros after the last number in a decimal and it will still be the same.

&

2.500 would be an equivalent decimal because you can add infinity zeros after the last number in a decimal and it will still be the same.

2.5, 2.500, 2.5000, 2.50000, etc..

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PLEASE HELP IN ALGEBRA

Aleksandr [31]

Sound, earthquakes, the brightness of stars, and chemistry, such as pH balance, a measure of acidity and alkalinity, are all instances of this.

7 0

3 years ago

Which number sentence is true?

Ilya [14]

The first one (when simplify) is

5.6>8.91

Which is not true if you plot it on a number line

The second one (when simplify) is

5.6<-8.91

which is not true because it is the same thing as number 1

The third one (when simplify) is

-5.6>8.91

which is obviously not true since negatives are way smaller than positives

So that leaves us the last one

Thefore the answer is OD.

I hope this helped!!

8 0

2 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!

navik [9.2K]

Answer:

Here, Exterior angles are ∠1, ∠2, ∠7 and ∠8

Interior angles are ∠3, ∠4, ∠5 and ∠6

Corresponding angles are ∠

(i) ∠1 and ∠5

(ii) ∠2 and ∠6

(iii) ∠4 and ∠8

(iv) ∠3 and ∠7

Axiom 4 If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.

Thus, (i) ∠1 = ∠5, (ii) ∠2 = ∠6, (iii) ∠4 = ∠8 and (iv) ∠3 = ∠7

Alternate Interior Angles: (i) ∠4 and ∠6 and (ii) ∠3 and ∠5

Alternate Exterior Angles: (i) ∠1 and ∠7 and (ii) ∠2 and ∠8

If a transversal intersects two parallel lines then each pair of alternate interior and exterior angles are equal.

Alternate Interior Angles: (i) ∠4 = ∠6 and (ii) ∠3 = ∠5

Alternate Exterior Angles: (i) ∠1 = ∠7 and (ii) ∠2 = ∠8

Interior angles on the same side of the transversal line are called the consecutive interior angles or allied angles or co-interior angles. They are as follows: (i) ∠4 and ∠5, and (ii) ∠3 and ∠6

5 0

3 years ago

Jamie used the distributive property to find the product of s(t) and h(t). His work was marked incorrect. Identify Jamie's mista

SSSSS [86.1K]

Complete Question:

Jamie used the distributive property to find the product of s(t) and h(t). His work was marked incorrect. Identify Jamie's mistake. What advice would you give Jamie to avoid this mistake in the future?

s(t)•h(t)= (3x-4)(2x-8)

= 6x² - 24x -8x - 32

= 6x² - 32x - 32

Answer:

Jamie made a mistake in his second line (6x² - 24x -8x - 32), by wrongly multiplying the operation signs. The last term should be +32, not -32.

Advice: Jamie should take note of the rule that applies when multiplying signs.

Step-by-step Explanation::

To find out where exactly Jamie made mistake, let's find the product of the given functions, step by step:

s(t)•h(t)= (3x-4)(2x-8)

Using distributive property, do the following:

$3x(2x - 8) -4(2x - 8)$

$3x*2x -3x*8 -4*2x -4*-8$

$6x^2 - 24x - 8x + 32$ (this is where Jamie made mistake. -4 * -8 = +32. NOT -32.)

Add like terms

$6x^2 - 32x + 32$

Jamie made a mistake in multiplying negative × negative. The last term in "6x² - 24x -8x - 32", should be +32. Negative × negative = +.

Therefore, it is advisable for Jamie to always take note of the rule that applies when multiplying signs.

8 0

3 years ago

Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i

Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

$a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333$

So the correct answer is C, the area between the two functions is 4/3 units.

3 0

3 years ago