Ask question

Ask question

All categories

3 years ago

12

How do I find square roots without a calculator?

2 answers:

sasho [114]3 years ago

8 0

Shahshjfohogkdjejeokdkdkf

tekilochka [14]3 years ago

6 0

You have to find to numbers that are the same and multiply them to see if you the answer your looking for for example, square root of 36 is 6 because 6 time 6 is 36

You might be interested in

Suppose triangle ABC will be dilated using the rule . What will be the distance from the center of dilation, Q, to the image of

Blababa [14]

Answer:

4

Step-by-step explanation:

The rule of dilation is $D_{Q,\frac{2}{3}}.$ This means that each distance QX' is equal to $\dfrac{2}{3}QX,$ where X is initial point and X' is its image after dilation.

Since QA=6, then

$QA'=\dfrac{2}{3}\cdot 6=4.$

6 0

2 years ago

Read 2 more answers

Amy and Jade bought packages of gum. Amy bought a total of 48 pieces. Jade bought a total of 32 pieces. If each package had the

dolphi86 [110]

Answer:

16 pieces were in each pack of gum

7 0

3 years ago

Arianna is preheating her oven before using it to bake. Arianna created the following

fredd [130]

Answer:

The change in degrees Fahrenheit for every change of one minute.

Step-by-step explanation:

7 0

3 years ago

A survey asked students at Wilson Academy which status they would prefer—to be happy, healthy, or rich—and how much pressure the

Brrunno [24]

Answer:

A. A student who prefers to be healthy is more likely to feel very little pressure from homework than a student who prefers to be rich.

Step-by-step explanation:

7 0

3 years ago

What is the area of the region bounded between the curves y=x and y=sqrt(x)?

erastova [34]

Answer:

$\frac{1}{6}$

Step-by-step explanation:

y = x .....(1)

$y=\sqrt{x}$ .... (2)

By solving equation (1) and equation (2)

$x = \sqrt{x}$

$\sqrt{x}\left ( \sqrt{x}-1 \right )=0$

$\sqrt{x}=0$ or $\left ( \sqrt{x}-1 \right )=0$

x = 0, x = 1

y = 0, y = 1

A = $\int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)$

A = $\int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx$

A = $\frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}$

A = $\frac{2}{3}-\frac{1}{2}$

A = $\frac{1}{6}$

8 0

3 years ago