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Anuta_ua [19.1K]

3 years ago

5

On Saturday, 15,598 people attended a music festival. How does the value represented by the 5 in the hundreds place compare with

the value of the 5 in the thousands place? *
a. The value in the hundreds place is 10 times as much as the value in the thousands place
b. The value in the hundreds place is one-tenth as much as the value in the thousands place
c. The value in the hundreds place is 100 times as much as the value in the thousands place.
d. The value in the hundreds place is one one-hundredth as much as the value in the thousands place

2 answers:

Vadim26 [7]3 years ago

6 0

Answer:

Gahsbdhhajauhdussjm

Step-by-step explanation:

Jajasjsjsjzjbdbsnsndjjz

ikadub [295]3 years ago

6 0

A because it is correct (not sure what else to say)

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James bought a cake that weighs 3 and 1 over 4 pounds. How many ounces does the cake weigh? Show your work. (5 points)

kolezko [41]

The weight of James cake in ounces is 52 ounces and the Pint of water dispensed in 20 seconds is 24 pints

<h3>Weight</h3>

Weight of cake = 3 1/4 pounds

1 pound = 16 ounces

Number of ounces the cake weigh = 3 1/4 pounds × 16 ounces

= 13/4 × 16

= (208) / 4

= 52 ounces

Water dispensed per second = 0.15 gallons
Number of seconds = 20 seconds

1 gallon = 4 quarts

0.15 gallon = 0.6 quarts

1 quart = 2 pints

0.6 quart = 1.2 pints

Pint of water dispensed in 20 seconds = 1.2 × 20

= 24 pints of water

Learn more about weight:

brainly.com/question/229459

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8 0

2 years ago

In a random sample of 85 people attending a workshop, 63 of the respondents said the workshop was useful. What is the sample pro

SVETLANKA909090 [29]

Divide the number of respondents saying it was useful by the total number
63/85 = 0.74

8 0

3 years ago

Read 2 more answers

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago

(See diagram) The gradient of the line joining the origin to the point A is 1/2. The distance between A and the origin is sqrt 2

GuDViN [60]

Answer:

(42, 21) or (–42, –21)

Step-by-step explanation:

let A(x, y), A on line 2y = x

distance

x² + y² = 2205

(2y)² + y² = 2205

y² = 441

y = ±21

x = ±42

3 0

4 years ago

Simplify the expression.<br><br> -2x + -5 + 8x + -4<br><br> Question 3 options:

Leokris [45]

6x + (-9) or 6x - 9

Try looking at the terms for the same things. Such as the variable like X and the negatives add together and make a negative

5 0

3 years ago