Ask question

Ask question

All categories

3 years ago

6

Estimate √50 to the hundredths place. 1. Estimate between two whole numbers: 72 = 49, 82 = 64 2. Estimate further to the tenths

place: 7.02 = 49.0, 7.12 = 50.41 3. Estimate further to the hundredths place:

2 answers:

Mumz [18]3 years ago

7 0

Answer:The √50 is between 7.07 and 7.08

Step-by-step explanation:

Gennadij [26K]3 years ago

5 0

Answer with explanation:

First Method

$\sqrt{50}=\sqrt{5\times 5 \times 2}\\\\=5\sqrt{2}\\\\=5 \times 1.41421\\\\=7.07105\\\\7.07$

Second method which is described in the question

$7 < \sqrt{50}$

You might be interested in

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

3 years ago

The length of the hypotenuse of a 30-60-90 triangle is 9. What is the perimeter?

denis23 [38]

The perimeter of any figure, is all its sides summed up

now, notice the picture below, using the 30-60-90 rule, that's "x"

now, get the other sides, since you know what "x" is, and sum them up, that's the perimeter

5 0

3 years ago

Read 2 more answers

A painter leans a 15 ft ladder against a building. The base of the ladder is 6 ft from the building. To the nearest foot, how hi

stiv31 [10]

Using the Pythagoras theorem

15^2 = x^2 + h^2 where h = height of ladder on the nuiding

h^2 = 15^2 - 6^2 = 189

= 13.75 ft to nearest hundredth

3 0

3 years ago

Read 2 more answers

Which statement can be used to prove that a given parallelogram is a rectangle?

RSB [31]

The <em>correct answer</em> is:

The diagonals of the parallelogram are congruent.

Explanation:

In every parallelogram, opposite angles are congruent. This would not mean it is a rectangle.

Consecutive sides of a parallelogram are only congruent if the parallelogram is a rhombus or a square; this would not be a rectangle.

The diagonals of every parallelogram bisect each other. This would not mean it is a rectangle.

The diagonals of a rectangle bisect each other. If we know this is true about our parallelogram, this means our parallelogram is a rectangle.

7 0

3 years ago

Read 2 more answers

The water tank in your school holds 20 liters of water. One day it was 34 full. That day, 14 liters of tank capacity was used up

Alex787 [66]

Answer:

20

Step-by-step explanation:

The water tank in the school hold 20 liters of water

One day it was 34 liters full

14 liters of water was used up

Therefore the quantity of water left can be calculated as follows

= 34-14

= 20

Hence 20 liters of water remains in the tank

6 0

3 years ago