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3 years ago

What is the value of the digit 7 in 870,541?

Mathematics

2 answers:

dem82 [27]3 years ago

7 0

10 thousand or 70,000

rosijanka [135]3 years ago

4 0

The value is 70,000 or seventy thousand

You might be interested in

Determine an equivalent algebraic monomial expression using the laws of exponents. Show your work.

Serga [27]

The equivalent algebraic monomial expression of the expression given as (-8a^5b)(3ab^4) is -24a^6b^5

<h3>How to determine an equivalent algebraic monomial expression?</h3>

The expression is given as:

(-8a^5b)(3ab^4)

Multiply -8 and 3

So, we have:

(-8a^5b)(3ab^4) = (-24a^5b)(ab^4)

Multiply a^5 and a (a^5 * a = a^6)

So, we have:

(-8a^5b)(3ab^4) = (-24a^6b)(b^4)

Multiply b and b^4

So, we have:

(-8a^5b)(3ab^4) = -24a^6b^5

Hence, the equivalent algebraic monomial expression of the expression given as (-8a^5b)(3ab^4) is -24a^6b^5

Read more about expressions at:

brainly.com/question/723406

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4 0

1 year ago

The cost of a plumbing repair is given by the equation y=45x+30, where x is the number of hours the repair takes and y is the to

arlik [135]

First you subtract the 30 from 300 and than 270/45 and your answer will be A. 6

6 0

3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

2 years ago

Determine the intercepts of the line.

aleksley [76]

Answer:

x-intercept : (-250, 0)

y-intercept : (0,100)

Step-by-step explanation:

Looking at the line plotted, you can see when and where the blue line intersects with the x-axis and y-axis.

6 0

3 years ago

What is the value of a in the figure shown?

Lemur [1.5K]

Answer:

Step-by-step explanation:

z = 180 - (180 - 114.8) - 45.6 = 114.8-45.6 = 69.2°

5 0

3 years ago