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2 years ago

Which of the following is a zero pair? a. 12, 12 b. -12, -12 C. 12, -12 d. O, 12

Mathematics

1 answer:

siniylev [52]2 years ago

3 0

Answer:I think it’s D

Step-by-step explanation:

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Which is the function and how do you know?

il63 [147K]

Answer:

Option 4

Step-by-step explanation:

3 0

2 years ago

Stephen evaluated (6.34 times 10 Superscript negative 7 Baseline) (4.5 times 10 Superscript 3 Baseline). His work is shown below

Maksim231197 [3]

Answer:

The step 4 and 5 describe the errors Stephen made.

Step-by-step explanation:

Given : Stephen evaluated $(6.34\times 10^{-7})(4.5\times 10^3)$ . His work is shown below.

To find : Which two statements describe the errors Stephen made?

Solution :

Step 1 - $(6.34\times 10^{-7})(4.5\times 10^3)$

Step 2 - $(6.34\times 4.5)(10^{-7}\times 10^3)$

Step 3 - $(28.53)(10^{-4})$

Step 4 - $28.53\times 10^{4}$

Error is the step as 4 sign is negative.

So correct step is $28.53\times 10^{-4}$

Step 5 - $28.53\times 10^{3}$

Error is the step as there is 4 in place of 3.

So correct step is $28.53\times 10^{-4}$

The step 4 and 5 describe the errors Stephen made.

7 0

3 years ago

Read 2 more answers

3. Which rule represents a linear function? Check all

Nady [450]

The answer is C

The answer is C because the function form for a linear function is f(x) = ax+b and the answer C has x which is the slope and -3 which would represent the y-intercept also known as b

7 0

3 years ago

Solve the equation for the indicated variable. SHOW ALL WORK! <br><br> ab-ac=2 solve for a

Juli2301 [7.4K]

Answer:

a= 2/b-c

Step-by-step explanation:

We want isolate for a or to be itself

ab-ac=2

factor out a

a (b-c)=2

Now bring b-c to the other side

a (b-c)/(b-c)=2/b-c

The b-c cancel each other out

a= 2/(b-c)

5 0

3 years ago

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

4 0

2 years ago