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2 years ago

Which fact is related to 63 divide by 7 =9?

1 answer:

8 0

That means that 7x9 is 63, or 63 divided by 7 is 9.” 3. OA. 7 Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations.

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THIS IS URGENT I NEED HELP ASAP.(9x 10^-3)^2

joja [24]

(9x 10^-3)^2= $\frac{81x^{2}}{1000000}$ , so d 8.1x10^-6

Evaluate 1/2a^-4b^2 for a =-2 and b=4

$\frac{1}{2} (-2)^{-4} *4^{2}\\\\\frac{1}{2} *\frac{1}{(-2)^{4}} *4^{2}\\\\\frac{1}{2} *\frac{1}{2^{4}}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*16=\frac{1}{2}$

A number raised to a negative exponent is sometimes negative.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

8 0

2 years ago

What is the average rate of change of the function from x = 0 to x = 2?.

irina1246 [14]

The average rate of change of the function from x = 0 to x = 2 is -2

From the complete question (see attachment), we have the following ordered pair when x = 0 and x = 2

$(x,y) = \{(0,10)\ (2,6)\}$

The average rate of change is calculated using:

$\Delta x = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$\Delta x = \frac{6-10}{2-0}$

Evaluate common terms

$\Delta x = \frac{-4}{2}$

Divide -4 by 2

$\Delta x = -2$

Hence, the average rate of change of the function from x = 0 to x = 2 is -2

Read more about rate of change at:

brainly.com/question/8728504

5 0

1 year ago

HELP!

artcher [175]

Answer:

B.Obtuse

as shown in the pictures that you put the angel is obtuse

6 0

2 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

5 0

2 years ago

If there wee 160 houses built and 3/8 of them are sold than how many are left

kramer

There would be 60 houses left

4 0

2 years ago