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3 years ago

Can anyone solve for B?

Mathematics

1 answer:

Alexeev081 [22]3 years ago

6 0

Answer:

x= 89

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Step-by-step explanation:

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Find the rate of change given the following solutions: (6, 2) and (1, 12).

blagie [28]

Answer:

The rate of change is -2

Step-by-step explanation:

f(b)-f(a) 12-2 10

---------- = ------------ = -------- = -2

b-a 1-6 -5

7 0

3 years ago

For a resistor in a direct current circuit that does not vary its resistance, the power that a resistor must dissipate is direct

Mrrafil [7]

THe problem is basically telling us: $P=kV^2$

where P is the power disappated and V^2 is our voltage squared.

$\frac{1}{16}=k*14^2\implies\\ \frac{1}{16*196}=k \implies \\ \frac{1}{3136}=k$

So, for the second example to find the power we simply have to plug k and our voltage back in, so: $P=\frac{14^2*3^2}{14^2*6} \implies \\ P=\frac{9}{6}= \frac{3}{2}$

6 0

3 years ago

Stella rewrites −212+3.7 as 3.7−212. Which property did she use? A. Additive Identity Property B. Additive Inverse Property C. C

lyudmila [28]

Answer:

Option C.

Step-by-step explanation:

It is given that Stella rewrites −212+3.7 as 3.7−212.

We need to find which property did she use.

According to commutative property of addition,

$A+B=B+A$

Where, A and B are any real values.

In the given expression −212+3.7, A=-212 and B=3.7.

Using commutative property of addition, we get

$(-212)+3.7=3.7+(-212)$

$(-212)+3.7=3.7-212$

So, she used commutative property.

Therefore, the correct option is C.

5 0

3 years ago

At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr

Dennis_Churaev [7]

The initial investment = $250
annual simple interest rate of 3% = 0.03

Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250

At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ The formula that can be used to find the account’s balance at the beginning of year n is:

A(n) = 250 + (n-1)(0.03 • 250)
At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5

So, the correct option is D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50

4 0

3 years ago

Read 2 more answers

Expressions equivalent to 3^10

ASHA 777 [7]

Answer:

59049+0

59048.1+0.9

59048.01+0.99

...

Step-by-step explanation:

There's infinite.

7 0

3 years ago

Read 2 more answers

Can anyone solve for B?​

Can anyone solve for B?