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3 years ago

Which shows two expressions that are equivalent to (- 8) (- 12) (2)?

Mathematics

2 answers:

Lana71 [14]3 years ago

8 0

Answer:

D is correct.

Step-by-step explanation:

Keep in mind that a negative multiplied by a negative is positive, and that a postive times a negative is a negative.

With this in mind the first expression -8*-12*2 will be a postive 192.

A is incorrect because -92*2 will be negative.

B is incorrect because -1*192 will be negative.

C is incorrect because -96*2 will be negative.

bonufazy [111]3 years ago

6 0

Answer: D (-8)(-24) and (-16)(-12)

Step-by-step explanation:

The first expression is equivalent to 192

Therefore, both must be positive which eliminates answer A, B, and C

Therefore the answer must be D

8x24 = 192

16x12=192

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Sarah bought a bike that cost $260. she had a coupon that was worth $55 off the cost of any bike. use the expression 260 + (-55)

mel-nik [20]

Answer:

Well, you can't add $55 on a coupon or it would cost more.

Step-by-step explanation:

Just subtract 260 by 55.

260 - 55 = $205 for the bike.

5 0

3 years ago

The area of a rectangle is 30 square root 3750 square inches, and the length is 3 square root 250 inches what is the width of th

Ann [662]

Answer:

$width=10\sqrt{15}\ in$

Step-by-step explanation:

-A rectangle's area is given by:

$A=lw\\\\l=length\\w=width$

Given A= $30\sqrt{3750}$ and $l=3\sqrt{250}$ , we substitute in the formula to solve for the width as follows:

$A=lw\\\\30\sqrt{3750}=3\sqrt{250}\times w\\\\w=\frac{30\sqrt{3750}}{3\sqrt{250}}\\\\w=10\frac{\sqrt{3750}}{\sqrt{250}}\\\\=10\times \sqrt{\frac{3750}{250}}\\\\=10\sqrt{15}\ in$

Hence , the rectangle's width is $10\sqrt{15}\ in$

6 0

3 years ago

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .