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2 years ago

Explain a way to find each sum 8 + 4

Mathematics

1 answer:

amm18122 years ago

6 0

Since we are adding two positive numbers here, their sum could be found by writing either

8
+4
---
12

or

4
+8
---
12

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Hester sells televisions. She earns a fixed amount for each television and an additional $ 15 if the buyer gets an extended war

tiny-mole [99]

Answer:

$y =mx +bx$

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

$1200 = 15*16 +16b$

and solving for b we got:

$b= \frac{1200-15*16}{16}= 60$

And then we can conclude that she earns 60 for each TV

Step-by-step explanation:

For this case we can set a linear model like this:

$y =mx +bx$

Where y is the total amount earned, m the amount for each extended warranty and b the fixed cost and x the total amount of TVs

For this case the value of x = 16 since we have 16 Tvs with extended warranties and we can do this:

$1200 = 15*16 +16b$

and solving for b we got:

$b= \frac{1200-15*16}{16}= 60$

And then we can conclude that she earns 60 for each TV

8 0

3 years ago

Which statement is true?

Allushta [10]

Im pretty sure it is the first one

5 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

5 0

2 years ago

Which expressions are completely factored?

amm1812

The only expressions that are correctly factored are;

A) 16a⁵ - 20a³ = 4a³(4a² - 5)

B) 24a⁴ + 18 = 6(4a⁴ + 3)

C) 12a³ + 8a = 4a(3a² + 2)

D) 30a⁶ - 24a² = 3a²(10a⁴ - 8)

<h3>How to factorize equations?</h3>

1) 16a⁵ - 20a³

To factorize this, we will have to get out the common factor first. The common factor is 4a³. Thus, we now have;

4a³(4a² - 5)

2) 24a⁴ + 18

To factorize this, we will have to get out the common factor first. The common factor is 6. Thus, we now have;

6(4a⁴ + 3)

3) 12a³ + 8a

To factorize this, we will have to get out the common factor first. The common factor is 4a. Thus, we now have;

4a(3a² + 2)

4) 30a⁶ - 24a²

To factorize this, we will have to get out the common factor first. The common factor is 3a². Thus, we now have;

3a²(10a⁴ - 8)

Read more about factorization of equations at; brainly.com/question/723406

#SPJ1

6 0

1 year ago

Pls answer quick it's gr 9 academic math lol

SVETLANKA909090 [29]

Answer:

1. (3^3 + 3^2)^2 actually equals (27 + 9)^2

which is the first mistake

2. (27 + 9)^2 does not equal (3^5)2, so (36)^2 does not equal 3^7

3. 3^7 DOES NOT EQUAL 21

Step-by-step explanation:

when you add powered numbers together, it does not multiply it, as your example:

1. (3^3 + 3^2)^2 actually equals (27 + 9)^2

which is the first mistake

2. (27 + 9)^2 does not equal (3^5)2, so (36)^2 does not equal 3^7

3. 3^7 DOES NOT EQUAL 21

4 0

2 years ago

Read 2 more answers